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how to check a propriety using r studio


Square root of Chi-square distribution tends to $N(0,1)$Relationship between chi-squared and standard normal distributions.How to check $H_0$ hypothesis using Pearson's criteria?Bivariate Normal Distribution Problem vs MarginalsShow that $Y = sum_{i=1}^n Y_i$ is distributed as $chi _{2n}^2$.If I have that $X sim chi^2_{1}$ and $Y sim chi^2_{2}$ are independent, how can I show that $4XY sim Y^2$?Chi-square test to check sampled varianceNormal distributionHypothesis testing: mean comparisonShow $1 + z_{alpha/2}/sqrt{n} approx chi^2_{alpha/2} (2n)/(2n)$













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










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
    11 hours ago










  • $begingroup$
    @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
    9 hours ago










  • $begingroup$
    @angryavian you can use R to sample from the distributions in question and use that to do some hypothesis testing.
    $endgroup$
    – JJJ
    7 hours ago
















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










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
    11 hours ago










  • $begingroup$
    @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
    9 hours ago










  • $begingroup$
    @angryavian you can use R to sample from the distributions in question and use that to do some hypothesis testing.
    $endgroup$
    – JJJ
    7 hours ago














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










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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 11 hours ago









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
    11 hours ago










  • $begingroup$
    @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
    9 hours ago










  • $begingroup$
    @angryavian you can use R to sample from the distributions in question and use that to do some hypothesis testing.
    $endgroup$
    – JJJ
    7 hours ago














  • 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
    11 hours ago










  • $begingroup$
    @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
    9 hours ago










  • $begingroup$
    @angryavian you can use R to sample from the distributions in question and use that to do some hypothesis testing.
    $endgroup$
    – JJJ
    7 hours ago








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
11 hours ago




$begingroup$
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
11 hours ago












$begingroup$
@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
9 hours ago




$begingroup$
@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
9 hours ago












$begingroup$
@angryavian you can use R to sample from the distributions in question and use that to do some hypothesis testing.
$endgroup$
– JJJ
7 hours ago




$begingroup$
@angryavian you can use R to sample from the distributions in question and use that to do some hypothesis testing.
$endgroup$
– JJJ
7 hours ago










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 10 hours ago









            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 9 hours ago

























                    answered 10 hours ago









                    BruceETBruceET

                    36.2k71540




                    36.2k71540























                        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 11 hours ago









                            George DewhirstGeorge Dewhirst

                            7164




                            7164






























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