how to check a propriety using r studio












2












$begingroup$


I have to check that this propriety



$Z sim N(0,1)$ and $Usim chi ^{2}(10)$ then $ Z/sqrt{U/10} sim T(10)$



is true using r studio if anyone can help , much appreciate






probability statistics hypothesis-testing







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    What do you mean by "verify using R"? A programming language cannot rigorously verify this although it may produce evidence suggesting it is true. If you read the definition of a $t$-distribution, then your question follows almost immediately.
    $endgroup$
    – angryavian
    Apr 5 at 16:52










  • $begingroup$
    @angryavian : Maybe not in r studio (although I don't know all the capabilities of R), but there's a thing called computer-assisted proofs.
    $endgroup$
    – Raskolnikov
    Apr 5 at 19:11










  • $begingroup$
    @angryavian you can use R to sample from the distributions in question and use that to do some hypothesis testing.
    $endgroup$
    – JJJ
    Apr 5 at 21:07
















2












$begingroup$


I have to check that this propriety



$Z sim N(0,1)$ and $Usim chi ^{2}(10)$ then $ Z/sqrt{U/10} sim T(10)$



is true using r studio if anyone can help , much appreciate






probability statistics hypothesis-testing







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    What do you mean by "verify using R"? A programming language cannot rigorously verify this although it may produce evidence suggesting it is true. If you read the definition of a $t$-distribution, then your question follows almost immediately.
    $endgroup$
    – angryavian
    Apr 5 at 16:52










  • $begingroup$
    @angryavian : Maybe not in r studio (although I don't know all the capabilities of R), but there's a thing called computer-assisted proofs.
    $endgroup$
    – Raskolnikov
    Apr 5 at 19:11










  • $begingroup$
    @angryavian you can use R to sample from the distributions in question and use that to do some hypothesis testing.
    $endgroup$
    – JJJ
    Apr 5 at 21:07














2












2








2


1



$begingroup$


I have to check that this propriety



$Z sim N(0,1)$ and $Usim chi ^{2}(10)$ then $ Z/sqrt{U/10} sim T(10)$



is true using r studio if anyone can help , much appreciate






probability statistics hypothesis-testing







share|cite|improve this question









$endgroup$




I have to check that this propriety



$Z sim N(0,1)$ and $Usim chi ^{2}(10)$ then $ Z/sqrt{U/10} sim T(10)$



is true using r studio if anyone can help , much appreciate






probability statistics hypothesis-testing




probability statistics hypothesis-testing






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Apr 5 at 16:46









JoshuaKJoshuaK

305




305








  • 3




    $begingroup$
    What do you mean by "verify using R"? A programming language cannot rigorously verify this although it may produce evidence suggesting it is true. If you read the definition of a $t$-distribution, then your question follows almost immediately.
    $endgroup$
    – angryavian
    Apr 5 at 16:52










  • $begingroup$
    @angryavian : Maybe not in r studio (although I don't know all the capabilities of R), but there's a thing called computer-assisted proofs.
    $endgroup$
    – Raskolnikov
    Apr 5 at 19:11










  • $begingroup$
    @angryavian you can use R to sample from the distributions in question and use that to do some hypothesis testing.
    $endgroup$
    – JJJ
    Apr 5 at 21:07














  • 3




    $begingroup$
    What do you mean by "verify using R"? A programming language cannot rigorously verify this although it may produce evidence suggesting it is true. If you read the definition of a $t$-distribution, then your question follows almost immediately.
    $endgroup$
    – angryavian
    Apr 5 at 16:52










  • $begingroup$
    @angryavian : Maybe not in r studio (although I don't know all the capabilities of R), but there's a thing called computer-assisted proofs.
    $endgroup$
    – Raskolnikov
    Apr 5 at 19:11










  • $begingroup$
    @angryavian you can use R to sample from the distributions in question and use that to do some hypothesis testing.
    $endgroup$
    – JJJ
    Apr 5 at 21:07








3




3




$begingroup$
What do you mean by "verify using R"? A programming language cannot rigorously verify this although it may produce evidence suggesting it is true. If you read the definition of a $t$-distribution, then your question follows almost immediately.
$endgroup$
– angryavian
Apr 5 at 16:52




$begingroup$
What do you mean by "verify using R"? A programming language cannot rigorously verify this although it may produce evidence suggesting it is true. If you read the definition of a $t$-distribution, then your question follows almost immediately.
$endgroup$
– angryavian
Apr 5 at 16:52












$begingroup$
@angryavian : Maybe not in r studio (although I don't know all the capabilities of R), but there's a thing called computer-assisted proofs.
$endgroup$
– Raskolnikov
Apr 5 at 19:11




$begingroup$
@angryavian : Maybe not in r studio (although I don't know all the capabilities of R), but there's a thing called computer-assisted proofs.
$endgroup$
– Raskolnikov
Apr 5 at 19:11












$begingroup$
@angryavian you can use R to sample from the distributions in question and use that to do some hypothesis testing.
$endgroup$
– JJJ
Apr 5 at 21:07




$begingroup$
@angryavian you can use R to sample from the distributions in question and use that to do some hypothesis testing.
$endgroup$
– JJJ
Apr 5 at 21:07










3 Answers
3






active

oldest

votes


















3












$begingroup$

One approach could be simulation of thousands of values:




  • Simulate $Z$ using rnorm

  • Simulate $U$ using rchisq

  • Do the division $Y = Z / sqrt{U / 10}$

  • Simulate the same number of $T$ from the hypothesised $t$-distribution using rt

  • Sort $Y$ and $T$ and plot them against each other - you want to see a diagonal straight line essentially $y=x$ with a little noise; this is visual demonstration though not a proof that the distributions are the same


You can do similar things with the qqplot function if you know what you are doing






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    I agree with @angryavian that you can't do a 'proof' in R.
    Also, it is crucial to state that random variables $Z$
    and $U$ are independent. Then $Y = frac{Z}{U/sqrt{10}} sim mathsf{T}(10)$ by definition.



    Here is R code to simulate a million values of $T$ [as in the Answer of @Henry (+1)], then to compare their histogram with the density of $mathsf{T}(10).$ This
    is a graphical demonstration that $T$ has (at least very nearly) the claimed t distribution.



    set.seed(405)  # for reproducibility
    
z = rnorm(10^6);  u = rchisq(10^6, 10)
    
y = z/sqrt(u/10)
    
hist(y, prob=T, br=50, col="skyblue2")
    
curve(dt(x, 10), add=T, lwd=2)


    enter image description here



    Furthermore, you could check that the quantiles of $Y$ very nearly match the theoretical quantiles of $mathsf{T}(10).$



    summary(y)
    
      Min.    1st Qu.     Median       Mean    3rd Qu.       Max. 
    
-10.641101  -0.699409   0.000059   0.000221   0.701253   9.802922 
    
qt(c(.25,.5,.75), 10)
    
[1] -0.6998121  0.0000000  0.6998121


    The summary above also shows that $bar Y approx 0.$ And the sample variance of the simulated values of $Y$ is very nearly the variance $nu/(nu - 2) = 10/8 = 1.25$ of Student's t distribution with $nu = 10$ degrees of freedom.
    [In effect, two of the moments suggested by #GeorgeDewhirts (+1).]



    var(y);  10/8
    
[1] 1.250115
    
[1] 1.25


    Also, you could do a Kolmogorov-Smirnov goodness-of-fit test on the first 5000 values of $Y$ and check that the P-value exceeds 5%. (The K-S test in R is limited to 5000 observations.) Roughly speaking, this is a formal, quantitative way to do @Henry's comparison of sorted observations.



    ks.test(y[1:5000], pt, 10)
    

    
    One-sample Kolmogorov-Smirnov test
    

    
data:  y[1:5000]
    
D = 0.013661, p-value = 0.3083
    
alternative hypothesis: two-sided





    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      You could compare the moments of your distribution with the theoretical moments of $T(10)$






      share|cite|improve this answer









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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        One approach could be simulation of thousands of values:




        • Simulate $Z$ using rnorm

        • Simulate $U$ using rchisq

        • Do the division $Y = Z / sqrt{U / 10}$

        • Simulate the same number of $T$ from the hypothesised $t$-distribution using rt

        • Sort $Y$ and $T$ and plot them against each other - you want to see a diagonal straight line essentially $y=x$ with a little noise; this is visual demonstration though not a proof that the distributions are the same


        You can do similar things with the qqplot function if you know what you are doing






        share|cite|improve this answer









        $endgroup$


















          3












          $begingroup$

          One approach could be simulation of thousands of values:




          • Simulate $Z$ using rnorm

          • Simulate $U$ using rchisq

          • Do the division $Y = Z / sqrt{U / 10}$

          • Simulate the same number of $T$ from the hypothesised $t$-distribution using rt

          • Sort $Y$ and $T$ and plot them against each other - you want to see a diagonal straight line essentially $y=x$ with a little noise; this is visual demonstration though not a proof that the distributions are the same


          You can do similar things with the qqplot function if you know what you are doing






          share|cite|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            One approach could be simulation of thousands of values:




            • Simulate $Z$ using rnorm

            • Simulate $U$ using rchisq

            • Do the division $Y = Z / sqrt{U / 10}$

            • Simulate the same number of $T$ from the hypothesised $t$-distribution using rt

            • Sort $Y$ and $T$ and plot them against each other - you want to see a diagonal straight line essentially $y=x$ with a little noise; this is visual demonstration though not a proof that the distributions are the same


            You can do similar things with the qqplot function if you know what you are doing






            share|cite|improve this answer









            $endgroup$



            One approach could be simulation of thousands of values:




            • Simulate $Z$ using rnorm

            • Simulate $U$ using rchisq

            • Do the division $Y = Z / sqrt{U / 10}$

            • Simulate the same number of $T$ from the hypothesised $t$-distribution using rt

            • Sort $Y$ and $T$ and plot them against each other - you want to see a diagonal straight line essentially $y=x$ with a little noise; this is visual demonstration though not a proof that the distributions are the same


            You can do similar things with the qqplot function if you know what you are doing







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Apr 5 at 17:52









            HenryHenry

            101k482170




            101k482170























                2












                $begingroup$

                I agree with @angryavian that you can't do a 'proof' in R.
                Also, it is crucial to state that random variables $Z$
                and $U$ are independent. Then $Y = frac{Z}{U/sqrt{10}} sim mathsf{T}(10)$ by definition.



                Here is R code to simulate a million values of $T$ [as in the Answer of @Henry (+1)], then to compare their histogram with the density of $mathsf{T}(10).$ This
                is a graphical demonstration that $T$ has (at least very nearly) the claimed t distribution.



                set.seed(405)  # for reproducibility
                
z = rnorm(10^6);  u = rchisq(10^6, 10)
                
y = z/sqrt(u/10)
                
hist(y, prob=T, br=50, col="skyblue2")
                
curve(dt(x, 10), add=T, lwd=2)


                enter image description here



                Furthermore, you could check that the quantiles of $Y$ very nearly match the theoretical quantiles of $mathsf{T}(10).$



                summary(y)
                
      Min.    1st Qu.     Median       Mean    3rd Qu.       Max. 
                
-10.641101  -0.699409   0.000059   0.000221   0.701253   9.802922 
                
qt(c(.25,.5,.75), 10)
                
[1] -0.6998121  0.0000000  0.6998121


                The summary above also shows that $bar Y approx 0.$ And the sample variance of the simulated values of $Y$ is very nearly the variance $nu/(nu - 2) = 10/8 = 1.25$ of Student's t distribution with $nu = 10$ degrees of freedom.
                [In effect, two of the moments suggested by #GeorgeDewhirts (+1).]



                var(y);  10/8
                
[1] 1.250115
                
[1] 1.25


                Also, you could do a Kolmogorov-Smirnov goodness-of-fit test on the first 5000 values of $Y$ and check that the P-value exceeds 5%. (The K-S test in R is limited to 5000 observations.) Roughly speaking, this is a formal, quantitative way to do @Henry's comparison of sorted observations.



                ks.test(y[1:5000], pt, 10)
                

                
    One-sample Kolmogorov-Smirnov test
                

                
data:  y[1:5000]
                
D = 0.013661, p-value = 0.3083
                
alternative hypothesis: two-sided





                share|cite|improve this answer











                $endgroup$


















                  2












                  $begingroup$

                  I agree with @angryavian that you can't do a 'proof' in R.
                  Also, it is crucial to state that random variables $Z$
                  and $U$ are independent. Then $Y = frac{Z}{U/sqrt{10}} sim mathsf{T}(10)$ by definition.



                  Here is R code to simulate a million values of $T$ [as in the Answer of @Henry (+1)], then to compare their histogram with the density of $mathsf{T}(10).$ This
                  is a graphical demonstration that $T$ has (at least very nearly) the claimed t distribution.



                  set.seed(405)  # for reproducibility
                  
z = rnorm(10^6);  u = rchisq(10^6, 10)
                  
y = z/sqrt(u/10)
                  
hist(y, prob=T, br=50, col="skyblue2")
                  
curve(dt(x, 10), add=T, lwd=2)


                  enter image description here



                  Furthermore, you could check that the quantiles of $Y$ very nearly match the theoretical quantiles of $mathsf{T}(10).$



                  summary(y)
                  
      Min.    1st Qu.     Median       Mean    3rd Qu.       Max. 
                  
-10.641101  -0.699409   0.000059   0.000221   0.701253   9.802922 
                  
qt(c(.25,.5,.75), 10)
                  
[1] -0.6998121  0.0000000  0.6998121


                  The summary above also shows that $bar Y approx 0.$ And the sample variance of the simulated values of $Y$ is very nearly the variance $nu/(nu - 2) = 10/8 = 1.25$ of Student's t distribution with $nu = 10$ degrees of freedom.
                  [In effect, two of the moments suggested by #GeorgeDewhirts (+1).]



                  var(y);  10/8
                  
[1] 1.250115
                  
[1] 1.25


                  Also, you could do a Kolmogorov-Smirnov goodness-of-fit test on the first 5000 values of $Y$ and check that the P-value exceeds 5%. (The K-S test in R is limited to 5000 observations.) Roughly speaking, this is a formal, quantitative way to do @Henry's comparison of sorted observations.



                  ks.test(y[1:5000], pt, 10)
                  

                  
    One-sample Kolmogorov-Smirnov test
                  

                  
data:  y[1:5000]
                  
D = 0.013661, p-value = 0.3083
                  
alternative hypothesis: two-sided





                  share|cite|improve this answer











                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    I agree with @angryavian that you can't do a 'proof' in R.
                    Also, it is crucial to state that random variables $Z$
                    and $U$ are independent. Then $Y = frac{Z}{U/sqrt{10}} sim mathsf{T}(10)$ by definition.



                    Here is R code to simulate a million values of $T$ [as in the Answer of @Henry (+1)], then to compare their histogram with the density of $mathsf{T}(10).$ This
                    is a graphical demonstration that $T$ has (at least very nearly) the claimed t distribution.



                    set.seed(405)  # for reproducibility
                    
z = rnorm(10^6);  u = rchisq(10^6, 10)
                    
y = z/sqrt(u/10)
                    
hist(y, prob=T, br=50, col="skyblue2")
                    
curve(dt(x, 10), add=T, lwd=2)


                    enter image description here



                    Furthermore, you could check that the quantiles of $Y$ very nearly match the theoretical quantiles of $mathsf{T}(10).$



                    summary(y)
                    
      Min.    1st Qu.     Median       Mean    3rd Qu.       Max. 
                    
-10.641101  -0.699409   0.000059   0.000221   0.701253   9.802922 
                    
qt(c(.25,.5,.75), 10)
                    
[1] -0.6998121  0.0000000  0.6998121


                    The summary above also shows that $bar Y approx 0.$ And the sample variance of the simulated values of $Y$ is very nearly the variance $nu/(nu - 2) = 10/8 = 1.25$ of Student's t distribution with $nu = 10$ degrees of freedom.
                    [In effect, two of the moments suggested by #GeorgeDewhirts (+1).]



                    var(y);  10/8
                    
[1] 1.250115
                    
[1] 1.25


                    Also, you could do a Kolmogorov-Smirnov goodness-of-fit test on the first 5000 values of $Y$ and check that the P-value exceeds 5%. (The K-S test in R is limited to 5000 observations.) Roughly speaking, this is a formal, quantitative way to do @Henry's comparison of sorted observations.



                    ks.test(y[1:5000], pt, 10)
                    

                    
    One-sample Kolmogorov-Smirnov test
                    

                    
data:  y[1:5000]
                    
D = 0.013661, p-value = 0.3083
                    
alternative hypothesis: two-sided





                    share|cite|improve this answer











                    $endgroup$



                    I agree with @angryavian that you can't do a 'proof' in R.
                    Also, it is crucial to state that random variables $Z$
                    and $U$ are independent. Then $Y = frac{Z}{U/sqrt{10}} sim mathsf{T}(10)$ by definition.



                    Here is R code to simulate a million values of $T$ [as in the Answer of @Henry (+1)], then to compare their histogram with the density of $mathsf{T}(10).$ This
                    is a graphical demonstration that $T$ has (at least very nearly) the claimed t distribution.



                    set.seed(405)  # for reproducibility
                    
z = rnorm(10^6);  u = rchisq(10^6, 10)
                    
y = z/sqrt(u/10)
                    
hist(y, prob=T, br=50, col="skyblue2")
                    
curve(dt(x, 10), add=T, lwd=2)


                    enter image description here



                    Furthermore, you could check that the quantiles of $Y$ very nearly match the theoretical quantiles of $mathsf{T}(10).$



                    summary(y)
                    
      Min.    1st Qu.     Median       Mean    3rd Qu.       Max. 
                    
-10.641101  -0.699409   0.000059   0.000221   0.701253   9.802922 
                    
qt(c(.25,.5,.75), 10)
                    
[1] -0.6998121  0.0000000  0.6998121


                    The summary above also shows that $bar Y approx 0.$ And the sample variance of the simulated values of $Y$ is very nearly the variance $nu/(nu - 2) = 10/8 = 1.25$ of Student's t distribution with $nu = 10$ degrees of freedom.
                    [In effect, two of the moments suggested by #GeorgeDewhirts (+1).]



                    var(y);  10/8
                    
[1] 1.250115
                    
[1] 1.25


                    Also, you could do a Kolmogorov-Smirnov goodness-of-fit test on the first 5000 values of $Y$ and check that the P-value exceeds 5%. (The K-S test in R is limited to 5000 observations.) Roughly speaking, this is a formal, quantitative way to do @Henry's comparison of sorted observations.



                    ks.test(y[1:5000], pt, 10)
                    

                    
    One-sample Kolmogorov-Smirnov test
                    

                    
data:  y[1:5000]
                    
D = 0.013661, p-value = 0.3083
                    
alternative hypothesis: two-sided






                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Apr 5 at 19:17

























                    answered Apr 5 at 18:26









                    BruceETBruceET

                    36.3k71540




                    36.3k71540























                        1












                        $begingroup$

                        You could compare the moments of your distribution with the theoretical moments of $T(10)$






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          You could compare the moments of your distribution with the theoretical moments of $T(10)$






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            You could compare the moments of your distribution with the theoretical moments of $T(10)$






                            share|cite|improve this answer









                            $endgroup$



                            You could compare the moments of your distribution with the theoretical moments of $T(10)$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Apr 5 at 17:16









                            George DewhirstGeorge Dewhirst

                            7514




                            7514






























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                                數位音樂下載

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                                數位音樂下載 為非以「實體」方式來販賣音樂的相關產品，而使用數位格式如mp3、AAC等文件格式來進行銷售販賣的一种音乐产业形式。著名的數位音樂下載服務商有iTunes Store、RecoChoku、Google Play音乐和微软Zune等。数字音乐下载在二十一世纪初一度风靡全球， [1] [2] 使得传统唱片产业走向衰落，但其自身也在2010年代流媒体音乐服务如Apple Music、Spotify等的挤压下迅速失去市场份额。 [3] [4] 数字音乐下载有便于携带、传播迅速、制作简单成本低等优点，与流媒体音乐相比也能更好保障音乐人权益； [5] 但它同时很可能带来侵犯版权和滥用的情形， [6] 也给音乐产业带来了诸多不利影响。数字音乐下载为音乐人和乐迷提供了更加便捷直接的交流渠道，也一定程度上推动了网络音乐人的出现和独立音乐的迅速发展。 [7] 目录 1 历史 1.1 技术积淀（1987-2000） 1.2 出现与兴起（2000-2004） 1.3 唱片市场的颠覆者（2004-2007） 1.4 巅峰和衰落（2005-2013） 1.5 摇摇欲坠的现状与不可预知的明天（2014至今） 2 主要文件格式 2.1 PCM编码格式 2.1.1 WAV 2.1.2 APE 2.1.3 FLAC 2.2 苹果推出的无损文件格式 2.2.1 AIFF 2.2.2 ALAC 2.3 有损压缩格式 2.3.1 MP3 2.3.2 AAC 3 主要市场 3.1 iTunes Store 3.2 Google Play音乐商店 3.3 亚马逊音乐商店 3.4 部分其他音乐商店 4 影响与评价 4.1 对音乐产业的影响 4.2 对科技行业的影响 4.3 正面评价 4.4 负面评价 4.4.1 侵权问题 4.4.2 经营模式不明晰 4.4.3 不适应移动时代 4.5 文化现象 5 延伸阅读 6 参考资料 历史 苹果iTunes+iPod的软硬件搭配让数字音
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                                格利澤436b

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                                格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP
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                                When can things happen in Etherscan, such as the picture below?

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                                2 I don't know why balance of the address is 0. when can this situation? etherscan balances share | improve this question asked yesterday Ru Hi Ru Hi 11 1 New contributor Ru Hi is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  | 
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