Limiting a sequence of moment generating functions












4














I was trying to solve the following problem:




Let ${X_n}_{n=1}^{infty}$ be a sequence of independent random variables with the probability mass function $P{X_n = pm1 } = frac{1}{2}$, $n in mathbb{N}$. Let $Z_n=sum_{j=1}^{n}{X_j/2^j}$. Show that $Z_n xrightarrow{L} Z$, where $Z sim U[-1, 1]$.




(From An Introduction to Probability and Statistics, V.K. Rohatgi & A. K. Md. Saleh, (c) 2015, Problems 7.5, Page 320)



Here $xrightarrow{L}$ means convergence in law (or in distribution), and $U[-1, 1]$ is the uniform distribution on the interval $[-1, 1]$.



My approach was the following:



We need to show that $$lim_{nrightarrowinfty} M_{Z_n}(t) = M_{Z}(t) = frac{e^{1 times t} - e^{-1 times t}}{t times (1 - (-1))} = frac{e^t - e^{-t}}{2t}.$$
Since
$$M_{Z_n}(t) = E_{Z_n}left(e^{tZ_n}right) = Eleft(e^{tsum_{j = 1}^{n}{frac{X_j}{2^j}}}right) = Eleft(prod_{j=1}^{n}{e^{tfrac{X_j}{2^j}}} right) = prod_{j=1}^{n}{E_{X_j}left(e^{tfrac{X_j}{2^j}} right)},\
E_{X_j}left( e^{t frac{X_j}{2^j}} right) = e^{t times frac{-1}{2^j}} times frac{1}{2} + e^{t times frac{1}{2^j}} times frac{1}{2} = frac{1}{2} left( e^{frac{t}{2^j}} + e^{frac{-t}{2^j}} right),$$

then
$$M_{Z_n}(t) = prod_{j = 1}^{n}{frac{1}{2} left( e^{frac{t}{2^j}} + e^{frac{-t}{2^j}} right)}.$$



I cannot see how this sequence of functions converges to the required moment generating function of $U[-1,1]$.



I had many attempts, for instance using the power series representation of $e^x$ and limiting approximations, but failed in them all. After that I started thinking that perhaps I am missing knowledge of some theorems.



Any idea how to proceed?










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  • Why not using characteristic functions?
    – Saad
    Dec 24 at 15:36










  • @Saad I tried that, but how does it make any difference? I thought that we use the characteristic function only when the moment generating function doesn't exist. However, here the moment generating function exists and is nice.
    – Noor AlYaqeen
    Dec 24 at 15:49


















4














I was trying to solve the following problem:




Let ${X_n}_{n=1}^{infty}$ be a sequence of independent random variables with the probability mass function $P{X_n = pm1 } = frac{1}{2}$, $n in mathbb{N}$. Let $Z_n=sum_{j=1}^{n}{X_j/2^j}$. Show that $Z_n xrightarrow{L} Z$, where $Z sim U[-1, 1]$.




(From An Introduction to Probability and Statistics, V.K. Rohatgi & A. K. Md. Saleh, (c) 2015, Problems 7.5, Page 320)



Here $xrightarrow{L}$ means convergence in law (or in distribution), and $U[-1, 1]$ is the uniform distribution on the interval $[-1, 1]$.



My approach was the following:



We need to show that $$lim_{nrightarrowinfty} M_{Z_n}(t) = M_{Z}(t) = frac{e^{1 times t} - e^{-1 times t}}{t times (1 - (-1))} = frac{e^t - e^{-t}}{2t}.$$
Since
$$M_{Z_n}(t) = E_{Z_n}left(e^{tZ_n}right) = Eleft(e^{tsum_{j = 1}^{n}{frac{X_j}{2^j}}}right) = Eleft(prod_{j=1}^{n}{e^{tfrac{X_j}{2^j}}} right) = prod_{j=1}^{n}{E_{X_j}left(e^{tfrac{X_j}{2^j}} right)},\
E_{X_j}left( e^{t frac{X_j}{2^j}} right) = e^{t times frac{-1}{2^j}} times frac{1}{2} + e^{t times frac{1}{2^j}} times frac{1}{2} = frac{1}{2} left( e^{frac{t}{2^j}} + e^{frac{-t}{2^j}} right),$$

then
$$M_{Z_n}(t) = prod_{j = 1}^{n}{frac{1}{2} left( e^{frac{t}{2^j}} + e^{frac{-t}{2^j}} right)}.$$



I cannot see how this sequence of functions converges to the required moment generating function of $U[-1,1]$.



I had many attempts, for instance using the power series representation of $e^x$ and limiting approximations, but failed in them all. After that I started thinking that perhaps I am missing knowledge of some theorems.



Any idea how to proceed?










share|cite|improve this question
























  • Why not using characteristic functions?
    – Saad
    Dec 24 at 15:36










  • @Saad I tried that, but how does it make any difference? I thought that we use the characteristic function only when the moment generating function doesn't exist. However, here the moment generating function exists and is nice.
    – Noor AlYaqeen
    Dec 24 at 15:49
















4












4








4


1





I was trying to solve the following problem:




Let ${X_n}_{n=1}^{infty}$ be a sequence of independent random variables with the probability mass function $P{X_n = pm1 } = frac{1}{2}$, $n in mathbb{N}$. Let $Z_n=sum_{j=1}^{n}{X_j/2^j}$. Show that $Z_n xrightarrow{L} Z$, where $Z sim U[-1, 1]$.




(From An Introduction to Probability and Statistics, V.K. Rohatgi & A. K. Md. Saleh, (c) 2015, Problems 7.5, Page 320)



Here $xrightarrow{L}$ means convergence in law (or in distribution), and $U[-1, 1]$ is the uniform distribution on the interval $[-1, 1]$.



My approach was the following:



We need to show that $$lim_{nrightarrowinfty} M_{Z_n}(t) = M_{Z}(t) = frac{e^{1 times t} - e^{-1 times t}}{t times (1 - (-1))} = frac{e^t - e^{-t}}{2t}.$$
Since
$$M_{Z_n}(t) = E_{Z_n}left(e^{tZ_n}right) = Eleft(e^{tsum_{j = 1}^{n}{frac{X_j}{2^j}}}right) = Eleft(prod_{j=1}^{n}{e^{tfrac{X_j}{2^j}}} right) = prod_{j=1}^{n}{E_{X_j}left(e^{tfrac{X_j}{2^j}} right)},\
E_{X_j}left( e^{t frac{X_j}{2^j}} right) = e^{t times frac{-1}{2^j}} times frac{1}{2} + e^{t times frac{1}{2^j}} times frac{1}{2} = frac{1}{2} left( e^{frac{t}{2^j}} + e^{frac{-t}{2^j}} right),$$

then
$$M_{Z_n}(t) = prod_{j = 1}^{n}{frac{1}{2} left( e^{frac{t}{2^j}} + e^{frac{-t}{2^j}} right)}.$$



I cannot see how this sequence of functions converges to the required moment generating function of $U[-1,1]$.



I had many attempts, for instance using the power series representation of $e^x$ and limiting approximations, but failed in them all. After that I started thinking that perhaps I am missing knowledge of some theorems.



Any idea how to proceed?










share|cite|improve this question















I was trying to solve the following problem:




Let ${X_n}_{n=1}^{infty}$ be a sequence of independent random variables with the probability mass function $P{X_n = pm1 } = frac{1}{2}$, $n in mathbb{N}$. Let $Z_n=sum_{j=1}^{n}{X_j/2^j}$. Show that $Z_n xrightarrow{L} Z$, where $Z sim U[-1, 1]$.




(From An Introduction to Probability and Statistics, V.K. Rohatgi & A. K. Md. Saleh, (c) 2015, Problems 7.5, Page 320)



Here $xrightarrow{L}$ means convergence in law (or in distribution), and $U[-1, 1]$ is the uniform distribution on the interval $[-1, 1]$.



My approach was the following:



We need to show that $$lim_{nrightarrowinfty} M_{Z_n}(t) = M_{Z}(t) = frac{e^{1 times t} - e^{-1 times t}}{t times (1 - (-1))} = frac{e^t - e^{-t}}{2t}.$$
Since
$$M_{Z_n}(t) = E_{Z_n}left(e^{tZ_n}right) = Eleft(e^{tsum_{j = 1}^{n}{frac{X_j}{2^j}}}right) = Eleft(prod_{j=1}^{n}{e^{tfrac{X_j}{2^j}}} right) = prod_{j=1}^{n}{E_{X_j}left(e^{tfrac{X_j}{2^j}} right)},\
E_{X_j}left( e^{t frac{X_j}{2^j}} right) = e^{t times frac{-1}{2^j}} times frac{1}{2} + e^{t times frac{1}{2^j}} times frac{1}{2} = frac{1}{2} left( e^{frac{t}{2^j}} + e^{frac{-t}{2^j}} right),$$

then
$$M_{Z_n}(t) = prod_{j = 1}^{n}{frac{1}{2} left( e^{frac{t}{2^j}} + e^{frac{-t}{2^j}} right)}.$$



I cannot see how this sequence of functions converges to the required moment generating function of $U[-1,1]$.



I had many attempts, for instance using the power series representation of $e^x$ and limiting approximations, but failed in them all. After that I started thinking that perhaps I am missing knowledge of some theorems.



Any idea how to proceed?







probability-theory probability-limit-theorems






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edited Dec 24 at 16:11









Saad

19.7k92252




19.7k92252










asked Dec 24 at 15:02









Noor AlYaqeen

286




286












  • Why not using characteristic functions?
    – Saad
    Dec 24 at 15:36










  • @Saad I tried that, but how does it make any difference? I thought that we use the characteristic function only when the moment generating function doesn't exist. However, here the moment generating function exists and is nice.
    – Noor AlYaqeen
    Dec 24 at 15:49




















  • Why not using characteristic functions?
    – Saad
    Dec 24 at 15:36










  • @Saad I tried that, but how does it make any difference? I thought that we use the characteristic function only when the moment generating function doesn't exist. However, here the moment generating function exists and is nice.
    – Noor AlYaqeen
    Dec 24 at 15:49


















Why not using characteristic functions?
– Saad
Dec 24 at 15:36




Why not using characteristic functions?
– Saad
Dec 24 at 15:36












@Saad I tried that, but how does it make any difference? I thought that we use the characteristic function only when the moment generating function doesn't exist. However, here the moment generating function exists and is nice.
– Noor AlYaqeen
Dec 24 at 15:49






@Saad I tried that, but how does it make any difference? I thought that we use the characteristic function only when the moment generating function doesn't exist. However, here the moment generating function exists and is nice.
– Noor AlYaqeen
Dec 24 at 15:49












3 Answers
3






active

oldest

votes


















2














$defe{mathrm{e}}$Another method: Note that $a + b = dfrac{a^2 - b^2}{a - b}$ for $a ≠ b$, then$$
M_{Z_n}(t) = frac{1}{2^n} prod_{k = 1}^n left( expleft( frac{t}{2^k} right) + expleft( -frac{t}{2^k} right) right) = frac{1}{2^n} · frac{e^t - e^{-t}}{expleft( dfrac{t}{2^n} right) - expleft( -dfrac{t}{2^n} right)}.
$$

Because$$
lim_{n → ∞} frac{expleft( dfrac{t}{2^n} right) - expleft( -dfrac{t}{2^n} right)}{dfrac{t}{2^n} - left( -dfrac{t}{2^n} right)} = (e^x)'bigr|_{x = 0} = 1,
$$

then $limlimits_{n → ∞} M_{Z_n}(t) = dfrac{1}{2t} (e^t - e^{-t})$ and the result follows.






share|cite|improve this answer





















  • Thanks this is the best answer!
    – Noor AlYaqeen
    Dec 24 at 18:45



















2














In addition to characteristic functions, one may also approach the problem via binary representation of integers, which is not as short as the answers with characteristic functions, but is quite straightforward.



Indeed, rewrite
$$
Z_n = frac{1}{2^n} sumlimits_{j=1}^n 2^{n-j} X_j : = frac{1}{2^n}S_n.
$$

Define $Lambda_{+} = { 1leq j leq 2^n: X_j = 1 }$, and let $Lambda_-$ be the complement of $Lambda_+$ in $1leq jleq 2^n$. Then,
$$
S_n = sum_{jin Lambda_+} - sum_{j in Lambda_-} = sum_{jin Lambda_+} - left( 2^n - 1 - sumlimits_{jin Lambda_+} right) = 2sumlimits_{j in Lambda_+} 2^{n-j} - (2^n - 1) tag{1}.
$$

Thus, with $S_n$ we cover all integers from $-(2^n - 1) , ... 2^n - 1$ of the form $(1)$, which are precisely all the odd integers from $-(2^n - 1), ... (2^n - 1)$, $2^n$ in total. Thus, if $i in [-(2^n - 1), ..., 2^n - 1] $ is even then
$$
mathbb{P}left(Z_n = frac{i}{2^n}right) = 0 tag{2}
$$

and if $i $ is odd, then
$$
mathbb{P}left(Z_n = frac{i}{2^n}right) = 2^{-n}, tag{3}
$$

since there is a single choice of index set $Lambda_+$ in $(1)$, and hence $(3)$ follows in view of independence of ${X_j}$.



From $(2)$ and $(3)$ we see, by counting the number of odd integers, that for any integer $-2^{n-1} + 1 leq i leq 2^{n-1}$ one has
$$
mathbb{P}left(Z_n leq frac{2i - 1}{2^n} right) = frac{1}{2} + frac{i}{2^n}.
$$



It follows that the distribution function $F_n$ of $Z_n$ coincides, on odd dyadic rationals from $[-1,1]$ (and obviously everywhere on $(-infty, -1] cup [1,infty)$) with the distribution function $F$ of a random variable with the law $U[-1, 1]$. The density of dyadic rationals and right-continuity of cdf imply $F_n to F$ everywhere on $[-1,1]$, hence the claim.






share|cite|improve this answer





















  • Thank you, a very interesting solution! although, a bit complicated.
    – Noor AlYaqeen
    Dec 24 at 18:33



















1














$defi{mathrm{i}}defd{mathrm{d}}$Since $X_1, X_2, cdots$ are independent and $displaystyle Z_n = sumlimits_{k = 1}^n frac{X_k}{2^k}$, then$$
φ_{Z_n}(t) = prod_{k = 1}^n φ_{X_k}left( frac{t}{2^k} right) = prod_{k = 1}^n cosleft( frac{t}{2^k} right) = frac{sin t}{2^n sinleft( dfrac{t}{2^n} right)}, quad forall t in mathbb{R}^*
$$

which implies $displaystyle lim_{n → ∞} φ_{Z_n}(t) = frac{sin t}{t}$, and the limit is the characteristic function of $Z sim U(-1, 1)$. By the continuity theorem, $Z_n xrightarrow{mathrm{d}} Z$.






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  • I didn't know you could give more than one answer...
    – maridia
    Dec 24 at 17:00










  • @Saad what was the identity you used to derive $prod_{k=1}^{n}{cosleft( frac{t}{2^k} right)} = frac{sin t}{2^n sinleft(frac{t}{2^n} right)}$?
    – Noor AlYaqeen
    Dec 24 at 18:21












  • @maridia: After you've posted an answer, the "Your Answer" textbox goes away, but you get an "Add Another Answer" button in its place. If you click that button, you get a confirmation dialog, "Are you sure you want to add another answer? ¶ You could use the edit link to refine and improve your existing answer, instead." If you click 'OK', you get the "Your Answer" textbox back.
    – ruakh
    Dec 24 at 21:32










  • @NoorAlYaqeen Note that $cosθ=dfrac{sin2θ}{sinθ}$.
    – Saad
    Dec 25 at 0:02











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3 Answers
3






active

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3 Answers
3






active

oldest

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active

oldest

votes






active

oldest

votes









2














$defe{mathrm{e}}$Another method: Note that $a + b = dfrac{a^2 - b^2}{a - b}$ for $a ≠ b$, then$$
M_{Z_n}(t) = frac{1}{2^n} prod_{k = 1}^n left( expleft( frac{t}{2^k} right) + expleft( -frac{t}{2^k} right) right) = frac{1}{2^n} · frac{e^t - e^{-t}}{expleft( dfrac{t}{2^n} right) - expleft( -dfrac{t}{2^n} right)}.
$$

Because$$
lim_{n → ∞} frac{expleft( dfrac{t}{2^n} right) - expleft( -dfrac{t}{2^n} right)}{dfrac{t}{2^n} - left( -dfrac{t}{2^n} right)} = (e^x)'bigr|_{x = 0} = 1,
$$

then $limlimits_{n → ∞} M_{Z_n}(t) = dfrac{1}{2t} (e^t - e^{-t})$ and the result follows.






share|cite|improve this answer





















  • Thanks this is the best answer!
    – Noor AlYaqeen
    Dec 24 at 18:45
















2














$defe{mathrm{e}}$Another method: Note that $a + b = dfrac{a^2 - b^2}{a - b}$ for $a ≠ b$, then$$
M_{Z_n}(t) = frac{1}{2^n} prod_{k = 1}^n left( expleft( frac{t}{2^k} right) + expleft( -frac{t}{2^k} right) right) = frac{1}{2^n} · frac{e^t - e^{-t}}{expleft( dfrac{t}{2^n} right) - expleft( -dfrac{t}{2^n} right)}.
$$

Because$$
lim_{n → ∞} frac{expleft( dfrac{t}{2^n} right) - expleft( -dfrac{t}{2^n} right)}{dfrac{t}{2^n} - left( -dfrac{t}{2^n} right)} = (e^x)'bigr|_{x = 0} = 1,
$$

then $limlimits_{n → ∞} M_{Z_n}(t) = dfrac{1}{2t} (e^t - e^{-t})$ and the result follows.






share|cite|improve this answer





















  • Thanks this is the best answer!
    – Noor AlYaqeen
    Dec 24 at 18:45














2












2








2






$defe{mathrm{e}}$Another method: Note that $a + b = dfrac{a^2 - b^2}{a - b}$ for $a ≠ b$, then$$
M_{Z_n}(t) = frac{1}{2^n} prod_{k = 1}^n left( expleft( frac{t}{2^k} right) + expleft( -frac{t}{2^k} right) right) = frac{1}{2^n} · frac{e^t - e^{-t}}{expleft( dfrac{t}{2^n} right) - expleft( -dfrac{t}{2^n} right)}.
$$

Because$$
lim_{n → ∞} frac{expleft( dfrac{t}{2^n} right) - expleft( -dfrac{t}{2^n} right)}{dfrac{t}{2^n} - left( -dfrac{t}{2^n} right)} = (e^x)'bigr|_{x = 0} = 1,
$$

then $limlimits_{n → ∞} M_{Z_n}(t) = dfrac{1}{2t} (e^t - e^{-t})$ and the result follows.






share|cite|improve this answer












$defe{mathrm{e}}$Another method: Note that $a + b = dfrac{a^2 - b^2}{a - b}$ for $a ≠ b$, then$$
M_{Z_n}(t) = frac{1}{2^n} prod_{k = 1}^n left( expleft( frac{t}{2^k} right) + expleft( -frac{t}{2^k} right) right) = frac{1}{2^n} · frac{e^t - e^{-t}}{expleft( dfrac{t}{2^n} right) - expleft( -dfrac{t}{2^n} right)}.
$$

Because$$
lim_{n → ∞} frac{expleft( dfrac{t}{2^n} right) - expleft( -dfrac{t}{2^n} right)}{dfrac{t}{2^n} - left( -dfrac{t}{2^n} right)} = (e^x)'bigr|_{x = 0} = 1,
$$

then $limlimits_{n → ∞} M_{Z_n}(t) = dfrac{1}{2t} (e^t - e^{-t})$ and the result follows.







share|cite|improve this answer












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share|cite|improve this answer










answered Dec 24 at 16:30









Saad

19.7k92252




19.7k92252












  • Thanks this is the best answer!
    – Noor AlYaqeen
    Dec 24 at 18:45


















  • Thanks this is the best answer!
    – Noor AlYaqeen
    Dec 24 at 18:45
















Thanks this is the best answer!
– Noor AlYaqeen
Dec 24 at 18:45




Thanks this is the best answer!
– Noor AlYaqeen
Dec 24 at 18:45











2














In addition to characteristic functions, one may also approach the problem via binary representation of integers, which is not as short as the answers with characteristic functions, but is quite straightforward.



Indeed, rewrite
$$
Z_n = frac{1}{2^n} sumlimits_{j=1}^n 2^{n-j} X_j : = frac{1}{2^n}S_n.
$$

Define $Lambda_{+} = { 1leq j leq 2^n: X_j = 1 }$, and let $Lambda_-$ be the complement of $Lambda_+$ in $1leq jleq 2^n$. Then,
$$
S_n = sum_{jin Lambda_+} - sum_{j in Lambda_-} = sum_{jin Lambda_+} - left( 2^n - 1 - sumlimits_{jin Lambda_+} right) = 2sumlimits_{j in Lambda_+} 2^{n-j} - (2^n - 1) tag{1}.
$$

Thus, with $S_n$ we cover all integers from $-(2^n - 1) , ... 2^n - 1$ of the form $(1)$, which are precisely all the odd integers from $-(2^n - 1), ... (2^n - 1)$, $2^n$ in total. Thus, if $i in [-(2^n - 1), ..., 2^n - 1] $ is even then
$$
mathbb{P}left(Z_n = frac{i}{2^n}right) = 0 tag{2}
$$

and if $i $ is odd, then
$$
mathbb{P}left(Z_n = frac{i}{2^n}right) = 2^{-n}, tag{3}
$$

since there is a single choice of index set $Lambda_+$ in $(1)$, and hence $(3)$ follows in view of independence of ${X_j}$.



From $(2)$ and $(3)$ we see, by counting the number of odd integers, that for any integer $-2^{n-1} + 1 leq i leq 2^{n-1}$ one has
$$
mathbb{P}left(Z_n leq frac{2i - 1}{2^n} right) = frac{1}{2} + frac{i}{2^n}.
$$



It follows that the distribution function $F_n$ of $Z_n$ coincides, on odd dyadic rationals from $[-1,1]$ (and obviously everywhere on $(-infty, -1] cup [1,infty)$) with the distribution function $F$ of a random variable with the law $U[-1, 1]$. The density of dyadic rationals and right-continuity of cdf imply $F_n to F$ everywhere on $[-1,1]$, hence the claim.






share|cite|improve this answer





















  • Thank you, a very interesting solution! although, a bit complicated.
    – Noor AlYaqeen
    Dec 24 at 18:33
















2














In addition to characteristic functions, one may also approach the problem via binary representation of integers, which is not as short as the answers with characteristic functions, but is quite straightforward.



Indeed, rewrite
$$
Z_n = frac{1}{2^n} sumlimits_{j=1}^n 2^{n-j} X_j : = frac{1}{2^n}S_n.
$$

Define $Lambda_{+} = { 1leq j leq 2^n: X_j = 1 }$, and let $Lambda_-$ be the complement of $Lambda_+$ in $1leq jleq 2^n$. Then,
$$
S_n = sum_{jin Lambda_+} - sum_{j in Lambda_-} = sum_{jin Lambda_+} - left( 2^n - 1 - sumlimits_{jin Lambda_+} right) = 2sumlimits_{j in Lambda_+} 2^{n-j} - (2^n - 1) tag{1}.
$$

Thus, with $S_n$ we cover all integers from $-(2^n - 1) , ... 2^n - 1$ of the form $(1)$, which are precisely all the odd integers from $-(2^n - 1), ... (2^n - 1)$, $2^n$ in total. Thus, if $i in [-(2^n - 1), ..., 2^n - 1] $ is even then
$$
mathbb{P}left(Z_n = frac{i}{2^n}right) = 0 tag{2}
$$

and if $i $ is odd, then
$$
mathbb{P}left(Z_n = frac{i}{2^n}right) = 2^{-n}, tag{3}
$$

since there is a single choice of index set $Lambda_+$ in $(1)$, and hence $(3)$ follows in view of independence of ${X_j}$.



From $(2)$ and $(3)$ we see, by counting the number of odd integers, that for any integer $-2^{n-1} + 1 leq i leq 2^{n-1}$ one has
$$
mathbb{P}left(Z_n leq frac{2i - 1}{2^n} right) = frac{1}{2} + frac{i}{2^n}.
$$



It follows that the distribution function $F_n$ of $Z_n$ coincides, on odd dyadic rationals from $[-1,1]$ (and obviously everywhere on $(-infty, -1] cup [1,infty)$) with the distribution function $F$ of a random variable with the law $U[-1, 1]$. The density of dyadic rationals and right-continuity of cdf imply $F_n to F$ everywhere on $[-1,1]$, hence the claim.






share|cite|improve this answer





















  • Thank you, a very interesting solution! although, a bit complicated.
    – Noor AlYaqeen
    Dec 24 at 18:33














2












2








2






In addition to characteristic functions, one may also approach the problem via binary representation of integers, which is not as short as the answers with characteristic functions, but is quite straightforward.



Indeed, rewrite
$$
Z_n = frac{1}{2^n} sumlimits_{j=1}^n 2^{n-j} X_j : = frac{1}{2^n}S_n.
$$

Define $Lambda_{+} = { 1leq j leq 2^n: X_j = 1 }$, and let $Lambda_-$ be the complement of $Lambda_+$ in $1leq jleq 2^n$. Then,
$$
S_n = sum_{jin Lambda_+} - sum_{j in Lambda_-} = sum_{jin Lambda_+} - left( 2^n - 1 - sumlimits_{jin Lambda_+} right) = 2sumlimits_{j in Lambda_+} 2^{n-j} - (2^n - 1) tag{1}.
$$

Thus, with $S_n$ we cover all integers from $-(2^n - 1) , ... 2^n - 1$ of the form $(1)$, which are precisely all the odd integers from $-(2^n - 1), ... (2^n - 1)$, $2^n$ in total. Thus, if $i in [-(2^n - 1), ..., 2^n - 1] $ is even then
$$
mathbb{P}left(Z_n = frac{i}{2^n}right) = 0 tag{2}
$$

and if $i $ is odd, then
$$
mathbb{P}left(Z_n = frac{i}{2^n}right) = 2^{-n}, tag{3}
$$

since there is a single choice of index set $Lambda_+$ in $(1)$, and hence $(3)$ follows in view of independence of ${X_j}$.



From $(2)$ and $(3)$ we see, by counting the number of odd integers, that for any integer $-2^{n-1} + 1 leq i leq 2^{n-1}$ one has
$$
mathbb{P}left(Z_n leq frac{2i - 1}{2^n} right) = frac{1}{2} + frac{i}{2^n}.
$$



It follows that the distribution function $F_n$ of $Z_n$ coincides, on odd dyadic rationals from $[-1,1]$ (and obviously everywhere on $(-infty, -1] cup [1,infty)$) with the distribution function $F$ of a random variable with the law $U[-1, 1]$. The density of dyadic rationals and right-continuity of cdf imply $F_n to F$ everywhere on $[-1,1]$, hence the claim.






share|cite|improve this answer












In addition to characteristic functions, one may also approach the problem via binary representation of integers, which is not as short as the answers with characteristic functions, but is quite straightforward.



Indeed, rewrite
$$
Z_n = frac{1}{2^n} sumlimits_{j=1}^n 2^{n-j} X_j : = frac{1}{2^n}S_n.
$$

Define $Lambda_{+} = { 1leq j leq 2^n: X_j = 1 }$, and let $Lambda_-$ be the complement of $Lambda_+$ in $1leq jleq 2^n$. Then,
$$
S_n = sum_{jin Lambda_+} - sum_{j in Lambda_-} = sum_{jin Lambda_+} - left( 2^n - 1 - sumlimits_{jin Lambda_+} right) = 2sumlimits_{j in Lambda_+} 2^{n-j} - (2^n - 1) tag{1}.
$$

Thus, with $S_n$ we cover all integers from $-(2^n - 1) , ... 2^n - 1$ of the form $(1)$, which are precisely all the odd integers from $-(2^n - 1), ... (2^n - 1)$, $2^n$ in total. Thus, if $i in [-(2^n - 1), ..., 2^n - 1] $ is even then
$$
mathbb{P}left(Z_n = frac{i}{2^n}right) = 0 tag{2}
$$

and if $i $ is odd, then
$$
mathbb{P}left(Z_n = frac{i}{2^n}right) = 2^{-n}, tag{3}
$$

since there is a single choice of index set $Lambda_+$ in $(1)$, and hence $(3)$ follows in view of independence of ${X_j}$.



From $(2)$ and $(3)$ we see, by counting the number of odd integers, that for any integer $-2^{n-1} + 1 leq i leq 2^{n-1}$ one has
$$
mathbb{P}left(Z_n leq frac{2i - 1}{2^n} right) = frac{1}{2} + frac{i}{2^n}.
$$



It follows that the distribution function $F_n$ of $Z_n$ coincides, on odd dyadic rationals from $[-1,1]$ (and obviously everywhere on $(-infty, -1] cup [1,infty)$) with the distribution function $F$ of a random variable with the law $U[-1, 1]$. The density of dyadic rationals and right-continuity of cdf imply $F_n to F$ everywhere on $[-1,1]$, hence the claim.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 24 at 17:04









Hayk

2,0721213




2,0721213












  • Thank you, a very interesting solution! although, a bit complicated.
    – Noor AlYaqeen
    Dec 24 at 18:33


















  • Thank you, a very interesting solution! although, a bit complicated.
    – Noor AlYaqeen
    Dec 24 at 18:33
















Thank you, a very interesting solution! although, a bit complicated.
– Noor AlYaqeen
Dec 24 at 18:33




Thank you, a very interesting solution! although, a bit complicated.
– Noor AlYaqeen
Dec 24 at 18:33











1














$defi{mathrm{i}}defd{mathrm{d}}$Since $X_1, X_2, cdots$ are independent and $displaystyle Z_n = sumlimits_{k = 1}^n frac{X_k}{2^k}$, then$$
φ_{Z_n}(t) = prod_{k = 1}^n φ_{X_k}left( frac{t}{2^k} right) = prod_{k = 1}^n cosleft( frac{t}{2^k} right) = frac{sin t}{2^n sinleft( dfrac{t}{2^n} right)}, quad forall t in mathbb{R}^*
$$

which implies $displaystyle lim_{n → ∞} φ_{Z_n}(t) = frac{sin t}{t}$, and the limit is the characteristic function of $Z sim U(-1, 1)$. By the continuity theorem, $Z_n xrightarrow{mathrm{d}} Z$.






share|cite|improve this answer





















  • I didn't know you could give more than one answer...
    – maridia
    Dec 24 at 17:00










  • @Saad what was the identity you used to derive $prod_{k=1}^{n}{cosleft( frac{t}{2^k} right)} = frac{sin t}{2^n sinleft(frac{t}{2^n} right)}$?
    – Noor AlYaqeen
    Dec 24 at 18:21












  • @maridia: After you've posted an answer, the "Your Answer" textbox goes away, but you get an "Add Another Answer" button in its place. If you click that button, you get a confirmation dialog, "Are you sure you want to add another answer? ¶ You could use the edit link to refine and improve your existing answer, instead." If you click 'OK', you get the "Your Answer" textbox back.
    – ruakh
    Dec 24 at 21:32










  • @NoorAlYaqeen Note that $cosθ=dfrac{sin2θ}{sinθ}$.
    – Saad
    Dec 25 at 0:02
















1














$defi{mathrm{i}}defd{mathrm{d}}$Since $X_1, X_2, cdots$ are independent and $displaystyle Z_n = sumlimits_{k = 1}^n frac{X_k}{2^k}$, then$$
φ_{Z_n}(t) = prod_{k = 1}^n φ_{X_k}left( frac{t}{2^k} right) = prod_{k = 1}^n cosleft( frac{t}{2^k} right) = frac{sin t}{2^n sinleft( dfrac{t}{2^n} right)}, quad forall t in mathbb{R}^*
$$

which implies $displaystyle lim_{n → ∞} φ_{Z_n}(t) = frac{sin t}{t}$, and the limit is the characteristic function of $Z sim U(-1, 1)$. By the continuity theorem, $Z_n xrightarrow{mathrm{d}} Z$.






share|cite|improve this answer





















  • I didn't know you could give more than one answer...
    – maridia
    Dec 24 at 17:00










  • @Saad what was the identity you used to derive $prod_{k=1}^{n}{cosleft( frac{t}{2^k} right)} = frac{sin t}{2^n sinleft(frac{t}{2^n} right)}$?
    – Noor AlYaqeen
    Dec 24 at 18:21












  • @maridia: After you've posted an answer, the "Your Answer" textbox goes away, but you get an "Add Another Answer" button in its place. If you click that button, you get a confirmation dialog, "Are you sure you want to add another answer? ¶ You could use the edit link to refine and improve your existing answer, instead." If you click 'OK', you get the "Your Answer" textbox back.
    – ruakh
    Dec 24 at 21:32










  • @NoorAlYaqeen Note that $cosθ=dfrac{sin2θ}{sinθ}$.
    – Saad
    Dec 25 at 0:02














1












1








1






$defi{mathrm{i}}defd{mathrm{d}}$Since $X_1, X_2, cdots$ are independent and $displaystyle Z_n = sumlimits_{k = 1}^n frac{X_k}{2^k}$, then$$
φ_{Z_n}(t) = prod_{k = 1}^n φ_{X_k}left( frac{t}{2^k} right) = prod_{k = 1}^n cosleft( frac{t}{2^k} right) = frac{sin t}{2^n sinleft( dfrac{t}{2^n} right)}, quad forall t in mathbb{R}^*
$$

which implies $displaystyle lim_{n → ∞} φ_{Z_n}(t) = frac{sin t}{t}$, and the limit is the characteristic function of $Z sim U(-1, 1)$. By the continuity theorem, $Z_n xrightarrow{mathrm{d}} Z$.






share|cite|improve this answer












$defi{mathrm{i}}defd{mathrm{d}}$Since $X_1, X_2, cdots$ are independent and $displaystyle Z_n = sumlimits_{k = 1}^n frac{X_k}{2^k}$, then$$
φ_{Z_n}(t) = prod_{k = 1}^n φ_{X_k}left( frac{t}{2^k} right) = prod_{k = 1}^n cosleft( frac{t}{2^k} right) = frac{sin t}{2^n sinleft( dfrac{t}{2^n} right)}, quad forall t in mathbb{R}^*
$$

which implies $displaystyle lim_{n → ∞} φ_{Z_n}(t) = frac{sin t}{t}$, and the limit is the characteristic function of $Z sim U(-1, 1)$. By the continuity theorem, $Z_n xrightarrow{mathrm{d}} Z$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 24 at 16:08









Saad

19.7k92252




19.7k92252












  • I didn't know you could give more than one answer...
    – maridia
    Dec 24 at 17:00










  • @Saad what was the identity you used to derive $prod_{k=1}^{n}{cosleft( frac{t}{2^k} right)} = frac{sin t}{2^n sinleft(frac{t}{2^n} right)}$?
    – Noor AlYaqeen
    Dec 24 at 18:21












  • @maridia: After you've posted an answer, the "Your Answer" textbox goes away, but you get an "Add Another Answer" button in its place. If you click that button, you get a confirmation dialog, "Are you sure you want to add another answer? ¶ You could use the edit link to refine and improve your existing answer, instead." If you click 'OK', you get the "Your Answer" textbox back.
    – ruakh
    Dec 24 at 21:32










  • @NoorAlYaqeen Note that $cosθ=dfrac{sin2θ}{sinθ}$.
    – Saad
    Dec 25 at 0:02


















  • I didn't know you could give more than one answer...
    – maridia
    Dec 24 at 17:00










  • @Saad what was the identity you used to derive $prod_{k=1}^{n}{cosleft( frac{t}{2^k} right)} = frac{sin t}{2^n sinleft(frac{t}{2^n} right)}$?
    – Noor AlYaqeen
    Dec 24 at 18:21












  • @maridia: After you've posted an answer, the "Your Answer" textbox goes away, but you get an "Add Another Answer" button in its place. If you click that button, you get a confirmation dialog, "Are you sure you want to add another answer? ¶ You could use the edit link to refine and improve your existing answer, instead." If you click 'OK', you get the "Your Answer" textbox back.
    – ruakh
    Dec 24 at 21:32










  • @NoorAlYaqeen Note that $cosθ=dfrac{sin2θ}{sinθ}$.
    – Saad
    Dec 25 at 0:02
















I didn't know you could give more than one answer...
– maridia
Dec 24 at 17:00




I didn't know you could give more than one answer...
– maridia
Dec 24 at 17:00












@Saad what was the identity you used to derive $prod_{k=1}^{n}{cosleft( frac{t}{2^k} right)} = frac{sin t}{2^n sinleft(frac{t}{2^n} right)}$?
– Noor AlYaqeen
Dec 24 at 18:21






@Saad what was the identity you used to derive $prod_{k=1}^{n}{cosleft( frac{t}{2^k} right)} = frac{sin t}{2^n sinleft(frac{t}{2^n} right)}$?
– Noor AlYaqeen
Dec 24 at 18:21














@maridia: After you've posted an answer, the "Your Answer" textbox goes away, but you get an "Add Another Answer" button in its place. If you click that button, you get a confirmation dialog, "Are you sure you want to add another answer? ¶ You could use the edit link to refine and improve your existing answer, instead." If you click 'OK', you get the "Your Answer" textbox back.
– ruakh
Dec 24 at 21:32




@maridia: After you've posted an answer, the "Your Answer" textbox goes away, but you get an "Add Another Answer" button in its place. If you click that button, you get a confirmation dialog, "Are you sure you want to add another answer? ¶ You could use the edit link to refine and improve your existing answer, instead." If you click 'OK', you get the "Your Answer" textbox back.
– ruakh
Dec 24 at 21:32












@NoorAlYaqeen Note that $cosθ=dfrac{sin2θ}{sinθ}$.
– Saad
Dec 25 at 0:02




@NoorAlYaqeen Note that $cosθ=dfrac{sin2θ}{sinθ}$.
– Saad
Dec 25 at 0:02


















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