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
    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
    8 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
    6 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






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
    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
    8 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
    6 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






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 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
    8 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
    6 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
    8 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
    6 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
8 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
8 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
6 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
6 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

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        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

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        active

        oldest

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        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 8 hours ago

























                    answered 9 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 10 hours ago









                            George DewhirstGeorge Dewhirst

                            7164




                            7164






























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                                Vallis Paradisi

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                                Coordinata: .mw-parser-output .geo-default,.mw-parser-output .geo-dms,.mw-parser-output .geo-dec{display:inline}.mw-parser-output .geo-nondefault,.mw-parser-output .geo-multi-punct{display:none}.mw-parser-output .longitude,.mw-parser-output .latitude{white-space:nowrap} 33°03′S 71°37′W  /  33.05°S 71.617°W  / -33.05; -71.617 Vallis Paradisi Res apud Vicidata repertae: Civitas : Chilia Numerus incolarum : 296 655 Zona horaria : UTC-4 Situs interretialis Nomen officiale : Valparaíso Gubernium Consilium : municipal council of Valparaíso Geographia Superficies : 401 600 000 Metrum quadratum Coniunctiones urbium Urbes gemellae : Ovetum , Badalona , Santa Fe , Ora longa , Sciamhaevum , Quantunia , Callao , Bat Yam , Melaka , Novorossiysk , Busan , Corduba , Odessa , Rosario , Malacca , Barcino , Soteropolis , Franciscopolis (California) , Guanajuato, Silao Tabula aut despectus Despectus in portum sinumque et urbem Vallis Paradisi Vallis Para...
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                                Tabula Rosettana

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                                Tabula multilinguis Rosettana in Museo Britannico ostenditur. Tabula Rosettana, [1] etiam titulo OGIS 90 agnita, est stela decreto de rebus sacris in Aegypto anno 196 a.C.n. lato inscripta. Tabula iuxta Rosettam Aegypti, urbem in delta Nili et ad oram maris Mediterranei iacentem, anno 1799 a milite Francico reperta est. Inventio stelae, linguis duabus scripturisque tribus inscriptae, eruditis Instituti Aegypti statim nuntiata est; ibi enim iussu imperatoris Napoleonis eruditi omnium scientiarum (sub aegide Commissionis Scientiarum et Artium) properaverant cum expeditione Francica. Qua a Britannis mox debellata, tabula Rosettana Londinium missa hodie apud Museum Britannicum iacet. Textus Graecus cito lectus interpretationi textuum Aegyptiorum (in formis hieroglyphica et demotica expressorum) gradatim adiuvit. Denique textum plene interpretatus est Ioannes Franciscus Champollion. Ab opere eruditorum cumulativo coepit hodiernus scripturae hieroglyphicae linguaeque Aegyptiae a...
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