Finding Approximate PDFs [closed]
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Any insight that can be given regarding this problem would be helpful! I'm not sure how to start this...
statistics
asked Dec 3 '18 at 19:49
Jason AJason A
6
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closed as off-topic by T. Bongers, Xander Henderson, Leucippus, KReiser, darij grinberg Dec 4 '18 at 2:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – T. Bongers, Xander Henderson, Leucippus, KReiser, darij grinberg
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Any insight that can be given regarding this problem would be helpful! I'm not sure how to start this...
statistics
asked Dec 3 '18 at 19:49
Jason AJason A
6
$endgroup$
closed as off-topic by T. Bongers, Xander Henderson, Leucippus, KReiser, darij grinberg Dec 4 '18 at 2:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – T. Bongers, Xander Henderson, Leucippus, KReiser, darij grinberg
If this question can be reworded to fit the rules in the help center, please edit the question.
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Generally, you will get better help here (without downvotes and votes to close) if you state the problem in your own words, and suggest possible approaches or show what you have tried and why you can't finish on your own.
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– BruceET
Dec 3 '18 at 22:24
add a comment |
$begingroup$
Any insight that can be given regarding this problem would be helpful! I'm not sure how to start this...
statistics
asked Dec 3 '18 at 19:49
Jason AJason A
6
$endgroup$
Any insight that can be given regarding this problem would be helpful! I'm not sure how to start this...
statistics
statistics
asked Dec 3 '18 at 19:49
Jason AJason A
6
asked Dec 3 '18 at 19:49
Jason AJason A
6
asked Dec 3 '18 at 19:49
Jason AJason A
6
asked Dec 3 '18 at 19:49
Jason AJason A
6
asked Dec 3 '18 at 19:49
Jason AJason A
6
6
closed as off-topic by T. Bongers, Xander Henderson, Leucippus, KReiser, darij grinberg Dec 4 '18 at 2:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – T. Bongers, Xander Henderson, Leucippus, KReiser, darij grinberg
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by T. Bongers, Xander Henderson, Leucippus, KReiser, darij grinberg Dec 4 '18 at 2:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – T. Bongers, Xander Henderson, Leucippus, KReiser, darij grinberg
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Generally, you will get better help here (without downvotes and votes to close) if you state the problem in your own words, and suggest possible approaches or show what you have tried and why you can't finish on your own.
$endgroup$
– BruceET
Dec 3 '18 at 22:24
add a comment |
$begingroup$
Generally, you will get better help here (without downvotes and votes to close) if you state the problem in your own words, and suggest possible approaches or show what you have tried and why you can't finish on your own.
$endgroup$
– BruceET
Dec 3 '18 at 22:24
$begingroup$
Generally, you will get better help here (without downvotes and votes to close) if you state the problem in your own words, and suggest possible approaches or show what you have tried and why you can't finish on your own.
$endgroup$
– BruceET
Dec 3 '18 at 22:24
$begingroup$
Generally, you will get better help here (without downvotes and votes to close) if you state the problem in your own words, and suggest possible approaches or show what you have tried and why you can't finish on your own.
$endgroup$
– BruceET
Dec 3 '18 at 22:24
add a comment |
2 Answers
2
active
oldest
votes
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So, you're asked to evaluate the Central Limit Theorem with this distribution at certain points.
My guess is that you should plug in the uniform distribution $mathcal{U}([0,1])$ and sum the values up. Given the CLT it should somewhat converge against the normal distribution $mathcal{N}(0,1)$ (at least I think so).
Be aware that the CLT assumes that all $X_i$ are iid.
answered Dec 3 '18 at 19:59
Thomas LangThomas Lang
1624
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$begingroup$
Not exactly an answer, but on the right track. (+1)
$endgroup$
– BruceET
Dec 3 '18 at 22:21
add a comment |
$begingroup$
Let $X_1, X_2, dots X_n$ be a random sample from
$mathsf{Unif}(0,1),$ with $mu = E(X_i) = 1/2$ and
$sigma^2 = Var(X_i) = 1/12.$
Then, according to the CLT the sample mean $bar X$ when $n = 25$ is approximately $mathsf{Norm}left(mu_n = 1/2,
sigma_n = sqrt{frac{1/12}{25}=0.5775}right).$
For $n = 25,$ the fit to normal is quite good, as shown in the figure below of a million simulated values of $bar X_{25}.$
For $n = 2$ the density function of $bar X_2$ has the shape
of an isosceles triangle with peak at 1/2, so the fit to normal is not so good.
The fit to normal from adding $n = 12$ standard uniform random variables is sufficiently close to normal that in early simulation programs (from
a time when
computation beyond $+$ and $-$ was very expensive) standard
normal random variables were simulated as
$Z = sum_{i=1}^{12} X_i - 6, = 12bar X_{12} - 6,$ where the $X_i$ are
standard uniform. With this formula it turns out that $E(Z) = 0$ and $Var(Z) = 1.$
The distribution of the sum of $n$ standard uniform distributions is called the Irwin-Hall distribution, illustrated in Wikipedia.
Note; R code for simulation in case it is of interest:
set.seed(1203); m = 10^6; n = 25
a.n = replicate(m, mean(runif(n)))
hist(a.n, prob=T, col="skyblue2",
main="Simulated Dist'n of Mean of 25 Std Unif RVs")
curve(dnorm(x, 1/2, .05774), add=T, lwd=2, col="red")
answered Dec 3 '18 at 21:44
BruceETBruceET
35.5k71440
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
So, you're asked to evaluate the Central Limit Theorem with this distribution at certain points.
My guess is that you should plug in the uniform distribution $mathcal{U}([0,1])$ and sum the values up. Given the CLT it should somewhat converge against the normal distribution $mathcal{N}(0,1)$ (at least I think so).
Be aware that the CLT assumes that all $X_i$ are iid.
answered Dec 3 '18 at 19:59
Thomas LangThomas Lang
1624
$endgroup$
$begingroup$
Not exactly an answer, but on the right track. (+1)
$endgroup$
– BruceET
Dec 3 '18 at 22:21
add a comment |
$begingroup$
So, you're asked to evaluate the Central Limit Theorem with this distribution at certain points.
My guess is that you should plug in the uniform distribution $mathcal{U}([0,1])$ and sum the values up. Given the CLT it should somewhat converge against the normal distribution $mathcal{N}(0,1)$ (at least I think so).
Be aware that the CLT assumes that all $X_i$ are iid.
answered Dec 3 '18 at 19:59
Thomas LangThomas Lang
1624
$endgroup$
$begingroup$
Not exactly an answer, but on the right track. (+1)
$endgroup$
– BruceET
Dec 3 '18 at 22:21
add a comment |
$begingroup$
So, you're asked to evaluate the Central Limit Theorem with this distribution at certain points.
My guess is that you should plug in the uniform distribution $mathcal{U}([0,1])$ and sum the values up. Given the CLT it should somewhat converge against the normal distribution $mathcal{N}(0,1)$ (at least I think so).
Be aware that the CLT assumes that all $X_i$ are iid.
answered Dec 3 '18 at 19:59
Thomas LangThomas Lang
1624
$endgroup$
So, you're asked to evaluate the Central Limit Theorem with this distribution at certain points.
My guess is that you should plug in the uniform distribution $mathcal{U}([0,1])$ and sum the values up. Given the CLT it should somewhat converge against the normal distribution $mathcal{N}(0,1)$ (at least I think so).
Be aware that the CLT assumes that all $X_i$ are iid.
answered Dec 3 '18 at 19:59
Thomas LangThomas Lang
1624
answered Dec 3 '18 at 19:59
Thomas LangThomas Lang
1624
answered Dec 3 '18 at 19:59
Thomas LangThomas Lang
1624
answered Dec 3 '18 at 19:59
Thomas LangThomas Lang
1624
1624
$begingroup$
Not exactly an answer, but on the right track. (+1)
$endgroup$
– BruceET
Dec 3 '18 at 22:21
add a comment |
$begingroup$
Not exactly an answer, but on the right track. (+1)
$endgroup$
– BruceET
Dec 3 '18 at 22:21
$begingroup$
Not exactly an answer, but on the right track. (+1)
$endgroup$
– BruceET
Dec 3 '18 at 22:21
$begingroup$
Not exactly an answer, but on the right track. (+1)
$endgroup$
– BruceET
Dec 3 '18 at 22:21
add a comment |
$begingroup$
Let $X_1, X_2, dots X_n$ be a random sample from
$mathsf{Unif}(0,1),$ with $mu = E(X_i) = 1/2$ and
$sigma^2 = Var(X_i) = 1/12.$
Then, according to the CLT the sample mean $bar X$ when $n = 25$ is approximately $mathsf{Norm}left(mu_n = 1/2,
sigma_n = sqrt{frac{1/12}{25}=0.5775}right).$
For $n = 25,$ the fit to normal is quite good, as shown in the figure below of a million simulated values of $bar X_{25}.$
For $n = 2$ the density function of $bar X_2$ has the shape
of an isosceles triangle with peak at 1/2, so the fit to normal is not so good.
The fit to normal from adding $n = 12$ standard uniform random variables is sufficiently close to normal that in early simulation programs (from
a time when
computation beyond $+$ and $-$ was very expensive) standard
normal random variables were simulated as
$Z = sum_{i=1}^{12} X_i - 6, = 12bar X_{12} - 6,$ where the $X_i$ are
standard uniform. With this formula it turns out that $E(Z) = 0$ and $Var(Z) = 1.$
The distribution of the sum of $n$ standard uniform distributions is called the Irwin-Hall distribution, illustrated in Wikipedia.
Note; R code for simulation in case it is of interest:
set.seed(1203); m = 10^6; n = 25
a.n = replicate(m, mean(runif(n)))
hist(a.n, prob=T, col="skyblue2",
main="Simulated Dist'n of Mean of 25 Std Unif RVs")
curve(dnorm(x, 1/2, .05774), add=T, lwd=2, col="red")
answered Dec 3 '18 at 21:44
BruceETBruceET
35.5k71440
$endgroup$
add a comment |
$begingroup$
Let $X_1, X_2, dots X_n$ be a random sample from
$mathsf{Unif}(0,1),$ with $mu = E(X_i) = 1/2$ and
$sigma^2 = Var(X_i) = 1/12.$
Then, according to the CLT the sample mean $bar X$ when $n = 25$ is approximately $mathsf{Norm}left(mu_n = 1/2,
sigma_n = sqrt{frac{1/12}{25}=0.5775}right).$
For $n = 25,$ the fit to normal is quite good, as shown in the figure below of a million simulated values of $bar X_{25}.$
For $n = 2$ the density function of $bar X_2$ has the shape
of an isosceles triangle with peak at 1/2, so the fit to normal is not so good.
The fit to normal from adding $n = 12$ standard uniform random variables is sufficiently close to normal that in early simulation programs (from
a time when
computation beyond $+$ and $-$ was very expensive) standard
normal random variables were simulated as
$Z = sum_{i=1}^{12} X_i - 6, = 12bar X_{12} - 6,$ where the $X_i$ are
standard uniform. With this formula it turns out that $E(Z) = 0$ and $Var(Z) = 1.$
The distribution of the sum of $n$ standard uniform distributions is called the Irwin-Hall distribution, illustrated in Wikipedia.
Note; R code for simulation in case it is of interest:
set.seed(1203); m = 10^6; n = 25
a.n = replicate(m, mean(runif(n)))
hist(a.n, prob=T, col="skyblue2",
main="Simulated Dist'n of Mean of 25 Std Unif RVs")
curve(dnorm(x, 1/2, .05774), add=T, lwd=2, col="red")
answered Dec 3 '18 at 21:44
BruceETBruceET
35.5k71440
$endgroup$
add a comment |
$begingroup$
Let $X_1, X_2, dots X_n$ be a random sample from
$mathsf{Unif}(0,1),$ with $mu = E(X_i) = 1/2$ and
$sigma^2 = Var(X_i) = 1/12.$
Then, according to the CLT the sample mean $bar X$ when $n = 25$ is approximately $mathsf{Norm}left(mu_n = 1/2,
sigma_n = sqrt{frac{1/12}{25}=0.5775}right).$
For $n = 25,$ the fit to normal is quite good, as shown in the figure below of a million simulated values of $bar X_{25}.$
For $n = 2$ the density function of $bar X_2$ has the shape
of an isosceles triangle with peak at 1/2, so the fit to normal is not so good.
The fit to normal from adding $n = 12$ standard uniform random variables is sufficiently close to normal that in early simulation programs (from
a time when
computation beyond $+$ and $-$ was very expensive) standard
normal random variables were simulated as
$Z = sum_{i=1}^{12} X_i - 6, = 12bar X_{12} - 6,$ where the $X_i$ are
standard uniform. With this formula it turns out that $E(Z) = 0$ and $Var(Z) = 1.$
The distribution of the sum of $n$ standard uniform distributions is called the Irwin-Hall distribution, illustrated in Wikipedia.
Note; R code for simulation in case it is of interest:
set.seed(1203); m = 10^6; n = 25
a.n = replicate(m, mean(runif(n)))
hist(a.n, prob=T, col="skyblue2",
main="Simulated Dist'n of Mean of 25 Std Unif RVs")
curve(dnorm(x, 1/2, .05774), add=T, lwd=2, col="red")
answered Dec 3 '18 at 21:44
BruceETBruceET
35.5k71440
$endgroup$
Let $X_1, X_2, dots X_n$ be a random sample from
$mathsf{Unif}(0,1),$ with $mu = E(X_i) = 1/2$ and
$sigma^2 = Var(X_i) = 1/12.$
Then, according to the CLT the sample mean $bar X$ when $n = 25$ is approximately $mathsf{Norm}left(mu_n = 1/2,
sigma_n = sqrt{frac{1/12}{25}=0.5775}right).$
For $n = 25,$ the fit to normal is quite good, as shown in the figure below of a million simulated values of $bar X_{25}.$
For $n = 2$ the density function of $bar X_2$ has the shape
of an isosceles triangle with peak at 1/2, so the fit to normal is not so good.
The fit to normal from adding $n = 12$ standard uniform random variables is sufficiently close to normal that in early simulation programs (from
a time when
computation beyond $+$ and $-$ was very expensive) standard
normal random variables were simulated as
$Z = sum_{i=1}^{12} X_i - 6, = 12bar X_{12} - 6,$ where the $X_i$ are
standard uniform. With this formula it turns out that $E(Z) = 0$ and $Var(Z) = 1.$
The distribution of the sum of $n$ standard uniform distributions is called the Irwin-Hall distribution, illustrated in Wikipedia.
Note; R code for simulation in case it is of interest:
set.seed(1203); m = 10^6; n = 25
a.n = replicate(m, mean(runif(n)))
hist(a.n, prob=T, col="skyblue2",
main="Simulated Dist'n of Mean of 25 Std Unif RVs")
curve(dnorm(x, 1/2, .05774), add=T, lwd=2, col="red")
answered Dec 3 '18 at 21:44
BruceETBruceET
35.5k71440
edited Dec 3 '18 at 22:29
answered Dec 3 '18 at 21:44
BruceETBruceET
35.5k71440
answered Dec 3 '18 at 21:44
BruceETBruceET
35.5k71440
answered Dec 3 '18 at 21:44
BruceETBruceET
35.5k71440
35.5k71440
add a comment |
add a comment |
$begingroup$
Generally, you will get better help here (without downvotes and votes to close) if you state the problem in your own words, and suggest possible approaches or show what you have tried and why you can't finish on your own.
$endgroup$
– BruceET
Dec 3 '18 at 22:24