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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 $1 + z_{alpha/2}/sqrt{n} approx chi^2_{alpha/2} (2n)/(2n)$
$begingroup$
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
asked 2 days ago
JoshuaKJoshuaK
305
$endgroup$
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
asked 2 days 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
2 days 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
2 days ago
$begingroup$
@angryavian you can use R to sample from the distributions in question and use that to do some hypothesis testing.
$endgroup$
– JJJ
2 days 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
asked 2 days ago
JoshuaKJoshuaK
305
$endgroup$
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
asked 2 days ago
JoshuaKJoshuaK
305
asked 2 days ago
JoshuaKJoshuaK
305
asked 2 days ago
JoshuaKJoshuaK
305
asked 2 days ago
JoshuaKJoshuaK
305
asked 2 days 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
2 days 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
2 days ago
$begingroup$
@angryavian you can use R to sample from the distributions in question and use that to do some hypothesis testing.
$endgroup$
– JJJ
2 days ago
add a comment |
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
2 days 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
2 days ago
$begingroup$
@angryavian you can use R to sample from the distributions in question and use that to do some hypothesis testing.
$endgroup$
– JJJ
2 days 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
2 days 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
2 days 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
2 days 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
2 days ago
$begingroup$
@angryavian you can use R to sample from the distributions in question and use that to do some hypothesis testing.
$endgroup$
– JJJ
2 days ago
$begingroup$
@angryavian you can use R to sample from the distributions in question and use that to do some hypothesis testing.
$endgroup$
– JJJ
2 days ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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
answered 2 days ago
HenryHenry
101k482170
$endgroup$
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
answered 2 days ago
BruceETBruceET
36.3k71540
$endgroup$
add a comment |
$begingroup$
You could compare the moments of your distribution with the theoretical moments of $T(10)$
answered 2 days ago
George DewhirstGeorge Dewhirst
7514
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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votes
active
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votes
$begingroup$
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
answered 2 days 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
answered 2 days 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
answered 2 days ago
HenryHenry
101k482170
$endgroup$
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
answered 2 days ago
HenryHenry
101k482170
answered 2 days ago
HenryHenry
101k482170
answered 2 days ago
HenryHenry
101k482170
answered 2 days ago
HenryHenry
101k482170
101k482170
add a comment |
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
answered 2 days ago
BruceETBruceET
36.3k71540
$endgroup$
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
answered 2 days ago
BruceETBruceET
36.3k71540
$endgroup$
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
answered 2 days ago
BruceETBruceET
36.3k71540
$endgroup$
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)
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
answered 2 days ago
BruceETBruceET
36.3k71540
edited 2 days ago
answered 2 days ago
BruceETBruceET
36.3k71540
answered 2 days ago
BruceETBruceET
36.3k71540
answered 2 days ago
BruceETBruceET
36.3k71540
36.3k71540
add a comment |
add a comment |
$begingroup$
You could compare the moments of your distribution with the theoretical moments of $T(10)$
answered 2 days ago
George DewhirstGeorge Dewhirst
7514
$endgroup$
add a comment |
$begingroup$
You could compare the moments of your distribution with the theoretical moments of $T(10)$
answered 2 days ago
George DewhirstGeorge Dewhirst
7514
$endgroup$
add a comment |
$begingroup$
You could compare the moments of your distribution with the theoretical moments of $T(10)$
answered 2 days ago
George DewhirstGeorge Dewhirst
7514
$endgroup$
You could compare the moments of your distribution with the theoretical moments of $T(10)$
answered 2 days ago
George DewhirstGeorge Dewhirst
7514
answered 2 days ago
George DewhirstGeorge Dewhirst
7514
answered 2 days ago
George DewhirstGeorge Dewhirst
7514
answered 2 days ago
George DewhirstGeorge Dewhirst
7514
7514
add a comment |
add a comment |
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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
2 days 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
2 days ago
$begingroup$
@angryavian you can use R to sample from the distributions in question and use that to do some hypothesis testing.
$endgroup$
– JJJ
2 days ago