Mrthfx

Why can't we play rap on piano?

Is it possible to run Internet Explorer on OS X El Capitan?

What's that red-plus icon near a text?

Has there ever been an airliner design involving reducing generator load by installing solar panels?

Character reincarnated...as a snail

Today is the Center

LWC SFDX source push error TypeError: LWC1009: decl.moveTo is not a function

"You are your self first supporter", a more proper way to say it

Why doesn't H₄O²⁺ exist?

Is it possible to do 50 km distance without any previous training?

Arrow those variables!

Why are electrically insulating heatsinks so rare? Is it just cost?

Languages that we cannot (dis)prove to be Context-Free

Replacing matching entries in one column of a file by another column from a different file

tikz convert color string to hex value

Fully-Firstable Anagram Sets

strTok function (thread safe, supports empty tokens, doesn't change string)

Can a Cauchy sequence converge for one metric while not converging for another?

How old can references or sources in a thesis be?

Intersection point of 2 lines defined by 2 points each

Approximately how much travel time was saved by the opening of the Suez Canal in 1869?

How to format long polynomial?

Why "Having chlorophyll without photosynthesis is actually very dangerous" and "like living with a bomb"?

How to determine what difficulty is right for the game?

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

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

asked 11 hours ago

JoshuaKJoshuaK

305

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

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

asked 11 hours ago

JoshuaKJoshuaK

305

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

$begingroup$

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

asked 11 hours ago

JoshuaKJoshuaK

305

$endgroup$

share|cite|improve this question

asked 11 hours ago

JoshuaKJoshuaK

305

share|cite|improve this question

asked 11 hours ago

JoshuaKJoshuaK

305

asked 11 hours ago

JoshuaKJoshuaK

305

asked 11 hours ago

JoshuaKJoshuaK

305

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

answered 10 hours ago

HenryHenry

101k482170

$endgroup$

add a comment | 

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)

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

edited 9 hours ago

answered 9 hours ago

BruceETBruceET

36.2k71540

$endgroup$

add a comment | 

1

$begingroup$

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

share|cite|improve this answer

answered 11 hours ago

George DewhirstGeorge Dewhirst

7164

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3176151%2fhow-to-check-a-propriety-using-r-studio%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

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

answered 10 hours ago

HenryHenry

101k482170

$endgroup$

add a comment | 

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

answered 10 hours ago

HenryHenry

101k482170

$endgroup$

add a comment | 

$begingroup$

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

answered 10 hours ago

HenryHenry

101k482170

$endgroup$

share|cite|improve this answer

answered 10 hours ago

HenryHenry

101k482170

answered 10 hours ago

HenryHenry

101k482170

answered 10 hours ago

HenryHenry

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)

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

edited 9 hours ago

answered 9 hours ago

BruceETBruceET

36.2k71540

$endgroup$

add a comment | 

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)

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

edited 9 hours ago

answered 9 hours ago

BruceETBruceET

36.2k71540

$endgroup$

add a comment | 

$begingroup$

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)

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

edited 9 hours ago

answered 9 hours ago

BruceETBruceET

36.2k71540

$endgroup$

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)

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

var(y);  10/8

[1] 1.250115

[1] 1.25

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

edited 9 hours ago

answered 9 hours ago

BruceETBruceET

36.2k71540

answered 9 hours ago

BruceETBruceET

36.2k71540

answered 9 hours ago

BruceETBruceET

36.2k71540

1

$begingroup$

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

share|cite|improve this answer

answered 11 hours ago

George DewhirstGeorge Dewhirst

7164

$endgroup$

add a comment | 

1

$begingroup$

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

share|cite|improve this answer

answered 11 hours ago

George DewhirstGeorge Dewhirst

7164

$endgroup$

add a comment | 

$begingroup$

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

share|cite|improve this answer

answered 11 hours ago

George DewhirstGeorge Dewhirst

7164

$endgroup$

share|cite|improve this answer

answered 11 hours ago

George DewhirstGeorge Dewhirst

7164

answered 11 hours ago

George DewhirstGeorge Dewhirst

7164

answered 11 hours ago

George DewhirstGeorge Dewhirst

7164