Expected Value of Order Statistics from Standard Normal Distribution

DHB10

New Member
Question: Find the expected value of the largest order statistic in a random sample of size 4 from the standard normal distribution.

Remarks: I got the only answer to this question without any derivation or proof. The definite integral is very difficult, please help with the integral. Please see the attached for details.

thanks a lot!

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Dragan

Super Moderator
Re: Expected Value of Order Statistics from Standard Normal Distribution--Need Help

Question: Find the expected value of the largest order statistic in a random sample of size 4 from the standard normal distribution.

Remarks: I got the only answer to this question without any derivation or proof. The definite integral is very difficult, please help with the integral. Please see the attached for details.

thanks a lot!

An easier way to handle this problem would be to fold the unit normal distribution at x=0 and integrate from 0 to Infinity. The approach that I am referring to is written in detail, for sample sizes of up to 7, in the following article (so sample size of 4 is included):

Renner, R.M (1976). Evaluation by power series of means of normal order statistics of samples of sizes six and seven. Mathematical Chronicle, vol. 4, no. 2-3, pp. 141-147.

DHB10

New Member
Hello, Dragan,
I am trying to use integral by parts and polar coordinates approach to handle it, however, I don't know how to determine the bounds at the end.

How can I get the article you mentioned above: Renner, R.M (1976). Evaluation by power series of means of normal order statistics of samples of sizes six and seven. Mathematical Chronicle, vol. 4, no. 2-3, pp. 141-147.

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DHB10

New Member
Integral Question When Evaluating Expected Value of Order Statistics

Question: Find the expected value of the largest order statistic in a random sample of size 4 from the standard normal distribution.

I am trying to use integral by parts and change to polar coordinaes to do the integral, but finially, I don't know how to determine the bounds. Please see attached for details.

Thanks

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Dragan

Super Moderator
Hello, Dragan,
I am trying to use integral by parts and polar coordinates approach to handle it, however, I don't know how to determine the bounds at the end.

How can I get the article you mentioned above: Renner, R.M (1976). Evaluation by power series of means of normal order statistics of samples of sizes six and seven. Mathematical Chronicle, vol. 4, no. 2-3, pp. 141-147.
Looks like you would have to order the article through a library to get it.

Give me a few minutes and I'll see if I can summarize the details to point you in the right direction.

And, you only need to make one thread on this topic.

Dragan

Super Moderator
Re: Integral Question When Evaluating Expected Value of Order Statistics

There's an easier way to do this. And your value of "a" in your first post isn't quite correct. I'll sketch it for you.

Let:

$$g\left ( x \right )=2f\left ( x \right )$$

$$G\left ( x \right )=2F\left ( x \right )-1$$

where $$f\left ( x \right )$$ and $$F\left ( x \right )$$ are the pdf and cdf for the standard normal distribution, so you only need to integrate over the positive real numbers as

$$I=\int_{0}^{\infty }\left [ G\left ( x \right ) \right ]^{3}g\left ( x \right )xdx$$

which can be integrated by parts as

$$I=3\int_{0}^{\infty }\left [ G\left ( x \right ) \right ]^{2}g\left ( x \right )^{2}dx$$

First:

$$\left [ G\left ( x \right ) \right ]^{2}=\frac{2}{\pi }\left \{ \int_{0}^{x} \exp \left ( -w^{2}/2 \right )dw\right \}^{2}$$

$$=\frac{2}{\pi }\int_{0}^{x}\int_{0}^{x}\exp \left ( -\frac{1}{2}\left ( x_{1}^{2}+x_{2}^{2} \right ) \right )dx_{1}dx_{2}$$

$$=\frac{4}{\pi }\int_{0}^{\frac{\pi }{4}}\left ( 1-\exp \left ( -\frac{1}{2} x^{2}\sec ^{2}\theta _{1}\right ) \right )d\theta _{1}$$

Now substitute and use Lichtenstien's theorem and integrate first with respect to x i.e.

$$\frac{2}{\pi }\int_{0}^{\infty }\exp \left ( -Ax^{2} \right )dx = \frac{1}{\sqrt{\pi A}}$$

where A is postive.

Thus, we have:

$$I=\frac{3}{\sqrt{\pi }}\left ( 1-\frac{4}{\pi } \int_{0}^{\frac{\pi }{4}}\frac{d\theta _{1}}{\sqrt{1+\frac{1}{2}\sec ^{2}\theta _{1}}}\right )$$

where

$$\int_{0}^{\frac{\pi }{4}}\frac{d\theta _{1}}{\sqrt{1+\frac{1}{2}\sec ^{2}\theta _{1}}} =\sin^{-1}\left ( \frac{1}{\sqrt{3}} \right )$$

which is what you're looking for.

DHB10

New Member
Re: Integral Question When Evaluating Expected Value of Order Statistics

Hello Dragan,

The answer in my first post is from an article which I searched and downloaded from Internet, it seems the article is part of a book. So, there is no author information. If you need the said article (PDF format), I can email to you

I double confirmed and verified that the answer is correct.

Today, I use the advice that you gave me when you first replied to my post, that is, fold the pdf and cdf. The attached is my detailed devrivation. but the final result is still different from the answer in my first post, which is correct, I have double confirmed and verified.

So, up to now, there are three answers to the same question, the first one is from an article from an unknown author, the second is from your above reply, the third is my today's derivation, which I used the idea that you told me : fold the pdf and cdf.

If the answer from the said unknown author is correct ( I am sure it is correct), where is our mistake in the derivation ? I double checked my today's devrivation, I didn't find out any mistake, please see the attached.

Other order statistic experts are more than welcomed to help with this question!

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Dragan

Super Moderator
Re: Integral Question When Evaluating Expected Value of Order Statistics

Well, I just checked an article that I wrote on this subject matter and I have:

$$\frac{3}{\sqrt{\pi }}-\frac{6\sin^{-1}\left ( \frac{1}{\sqrt{3}} \right )}{\pi ^{3/2}}=1.02937527...$$

which I believe to be correct.

DHB10

New Member
Re: Integral Question When Evaluating Expected Value of Order Statistics

Hello, Dragan,

Right now, I have known the reason why both your answer and my answer are different from the answer in my first post, which I had double confimed and verified its correctness. Please see the attached charts. Because the symmetrical axis of (pdf^2)*(cdf^2) is not Y axis. So, we need to find out the exact symmetrical axis of (pdf^2)*(cdf^2), and then "fold", do the integration.

The mistake in our previous post is that we wrongly took Y axis as the symmetrical axis of (pdf^2)*(cdf^2) , so, doing the integral from 0 to plus infinite is not correct.

I am trying to find out the exact symmetrical axis of (pdf^2)*(cdf^2), and then we can fold to go ahead doing the integral. I will post my finding. if you find first, please post your finding here. Thanks a lot in advance.

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DHB10

New Member
Re: Integral Question When Evaluating Expected Value of Order Statistics

Yes, 1.0293753730040....... is correct

DHB10

New Member
Re: Integral Question When Evaluating Expected Value of Order Statistics

Hello, Dragan,

Thanks

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Dragan

Super Moderator
Re: Integral Question When Evaluating Expected Value of Order Statistics

Hello, Dragan,

Thanks

For the first question, I am integrating by parts, that is:

$${G}'\left ( x \right )=g\left ( x \right )$$ and

$${g}'\left ( x \right )=-g\left ( x \right )x$$

Note that $$G\left ( 0 \right )=0$$ and $$\lim g\left ( x \right )=0$$ as $$x\rightarrow \infty$$

For the second part, $$G\left ( x \right )$$ is the c.d.f. associated with the 'folded normal' or chi-square distribution (one d.o.f).

$$G\left ( x \right )=2F\left ( x \right )-1$$ so that $$F\left ( x \right )=\frac{1}{2}+\frac{1}{2}\int_{0}^{\infty }g\left ( w \right )dw$$

And, no I am not changing to polar coordinates---that won't work. Rather, I am using the substitution $$x_{2}=x_{1}\tan\theta _{1}$$ and thus

$$dx_{2}=x_{1}\sec ^{2}\theta _{1}d\theta _{1}$$ Further, let $$x_{1}^{2}+x_{2}^{2}=x_{1}^{2}\sec ^{2}\theta _{1}$$. As such the region of integration will be reduced to one-half of the area of the origninal rectangle. The limits of the integral should be straight-forward.

DHB10

New Member
Re: Integral Question When Evaluating Expected Value of Order Statistics

Hello, Dragan,

The another answer in my first post, which is from an unknow author, is also equal to 1.02937527..... It seems that the two answers are the same or identical,

I have already proved that the two different explicit expressions of the answer are identical. Please see the attached for details.

It couldn't be better anymore if you would like to share your detailed derivation or proof and post it here.

Thanks a lot!

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DHB10

New Member
Re: Integral Question When Evaluating Expected Value of Order Statistics

Hello, Dragan,

I have three more questions need your help and clarification.

1.In your post dated on 01-22-2012 03:11 PM, there is an integral at the end of your post, I have already computed the integral, what I computed is differenct from yours. I double checked my computation, I failed to search out any mistake, So, it might be correct or incorrect. Please see the attached below for more and details. My first question is that: Are you 100% sure your computation of the integral is correct?

2.How did you derive or prove the expected value E(X(4)) in this topic is equal to the integral G(x)^3 * g(x)*x from 0 to infinite ? because according to the accepted moments definition, the expected value E(X(4)) in this topic should be equal to integral F(x)^3 *f(x) *x from minus infinite to plus infinite, where F(x) and f(x) are the cdf and pdf of the standard normal distribtion respectively.

In other words, how did you dervie or prove that E(X(4)) = integral G(x)^3*g(x)*x from 0 to plus infinite is the same as E(X(4)) = integral F(x)^3*f(x)*x from minus infinite to plus infinite ?

3.In your previous post, your G(x) is actually the Erf function of a standard normal distribution, or known as the error function in most of the statistic books, which should be G(x)= Erf(x) = 2/square root(PI) times integral exp(-w^2) from 0 to x. Why did you use square root(2/PI) instead of 2/square root(PI) ? I mean, Why your G(x)= squre root (2/PI) times integral (1/2) exp(-w^2) from 0 to x ? Please see the attached for my exact points of this question.

DHB10 from China, PRC

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Dragan

Super Moderator
Re: Integral Question When Evaluating Expected Value of Order Statistics

I'll take part one of your post for now because I'm busy. I'll get to the rest later. I think what you're missing is this:

$$\int_{0}^{\frac{\pi }{4}}\frac{d\theta _{1}}{\sqrt{1+\frac{1}{2}\sec ^{2}\theta _{1}}} =\int_{0}^{\frac{\pi }{4}}\frac{\cos \theta _{1}d\theta _{1}}{\sqrt{\frac{1}{2}+\cos ^{2}\theta _{1}}}=\int_{0}^{\alpha }du=\alpha =\sin^{-1}\left ( \frac{1}{\sqrt{3}} \right )$$

because the transformation $$\sin \theta =\sqrt{\frac{3}{2}}\sin u$$ can be applied to the left-hand side (i.e. the Jacobian of such a transformation is necessarily a diagonal form) and thus,
$$\cos ^{2}\theta +\frac{1}{2}=\frac{3}{2}\cos ^{2}u$$.

DHB10

New Member
Re: Integral Question When Evaluating Expected Value of Order Statistics

Hello, Dragan,

I have already mathematically proved that both your answer and my answer to the said integral are identical. both = 0.615479709...... For a rigorous mathematical proof, please see the attached below.

Best Regards
DHB10 from China, PRC

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DHB10

New Member
Re: Integral Question When Evaluating Expected Value of Order Statistics

Hello, Dragan,

I have already got the said article from the author, who sent it to me by email. So, Please ignore the remaining two questions in my previous post.

Thanks
DHB10 from China, PRC

DHB10

New Member
Re: Integral Question When Evaluating Expected Value of Order Statistics

Hello, Dragan,

I have already finished the rigorous mathematical proof of "folded" integral and full computation of this topic. To close this topic finally, I would like to share my detailed proof and full computation with you and those who might meet the same question or be concerned about this topic.

Please see the following attachments for details (4 pages in total).

Meanwhile, many thanks for your great help with the topic.

Best Regards
DHB10 from Shenzhen, China, PRC

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