Covariance of Order Statistics from N(0,1) Distribution

DHB10

New Member
#1
Hello,

I am verifing an exact value of the 2nd moment E(R^2) of a sample range (R) from the standard normal distribution. To let you clearly know what is the question that I am confronted with. Please see the attachments for details.

To save your time, you can only read the highlighted part on page 1 to get the context. My question is at the end of page 2.

Your help is highly appreciated !

Thanks a lot in advance.
DHB10 from China, PRC
 
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DHB10

New Member
#3
Hello,

My another related question:

Because we are dealing with the random variable XY, for sure, both X and Y follow the normal distribution respectively, Is the random variable Z= XY follows the bivariate normal distribution ? If yes, we can get the joint probability density funcction f(x,y) of the random variable XY as shown in equation (3) upstairs.

Thanks a lot in advance.
 
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Dason

Ambassador to the humans
#4
Are you talking about X= X(1) and Y=X(n)? Because in that case X and Y don't have normal distributions.
 

BGM

TS Contributor
#5
\( X_{(1)}, X_{(n)} \) are order statistics and in general they do not follow the same distribution as the original sample. So you cannot say they have a marginal/jointly normal distribution.

To obtain the joint pdf is not hard. For example, in your case you can follow the following arguments:

To obtain \( X_{(1)} = x_1, X_{(n)} = x_n \), it is equivalent to:

1. Having 0 sample smaller than \( x_1 \)
2. Having 1 sample at \( x_1 \)
3. Having n - 2 sample between \( x_1, x_n \)
4. Having 1 sample at \( x_n \)
5. Having 0 sample larger than \( x_n \)

And hence therefore using the multinomial arguments
\( f_{X_{(1)}, X_{(n)}}(x_1, x_n)dx_1dx_n \) \( = \frac {n!} {0!1!(n-2)!1!0!} F(x_1)^0 f(x_1)dx_1 [F(x_n) - F(x_1)]^{n-2} f(x_n)dx_n [1 - F(x_n)]^0 \)

\( \Rightarrow f_{X_{(1)}, X_{(n)}}(x_1, x_n) = n(n - 1)[F(x_n) - F(x_1)]^{n-2}f(x_1)f(x_n), -\infty < x_1 < x_n < +\infty \)
 

DHB10

New Member
#6
Hello, BGM & Dason,

Many thanks for your help.

To BGM, in your equation upstairs, are f(x) and F(x) the pdf and cdf of the standard normal distribution respectively?

Thanks
DHB10 from China, PRC
 

Dason

Ambassador to the humans
#7
The equation BGM posted gives the joint distribution for the min/max of a sample of size n for any distribution where F is the cdf and f is the pdf. In your case you would want to use the pdf and cdf of the standard normal.
 

DHB10

New Member
#8
Hello, Dason,

Many thanks for your explanation!

I am going to use both your way to compute the E(R^2).

DHB10 from CN
 
#10
Hello, Dragan,

Nice to hear you again and thanks for your information, I am going to see the article you mentioned upstairs.

Thanks again.
DHB10 from China.
 

BGM

TS Contributor
#11
The support of the joint pdf is \( -\infty < x_1 < x_n < +\infty \). So when you are integrating with this joint pdf with iterated integral, make sure that you get the integration limits of the inner integral correct - it is not integrating from \( - \infty \) to \( + \infty \) anymore.

For \( n = 3 \) example, you have missed the pdf after splitting the differences?
 
#12
Hello, BGM,

Many thanks for your help.

Meanwhile, I would like to show something interested in the following attachments. Although when n = 2, the computation result of E(R^2) is correct in my previous, at least, is the same as the author's version, however, it seems that it is coincidently correct, because according to some evidences, the probablity density function of the random variable Z = XY in this context is not what I used when computing E(R^2) in my previous post, it should have been the one shown in the following attachment ( page 3 of 3) based on the evidences shown in the following attachements.

DHB10 from China, PRC
 

BGM

TS Contributor
#14
You cannot substitute \( z = xy \). \( x \) is a dummy variable inside the integral which is independent of the (another dummy) variable \( z \)

Also, in this example you need to ensure that you are integrating within this set: \( -\infty < x < \frac {z} {x} < +\infty \), i.e. the integration limits are in terms of \( z \)

Lastly, it is not hard to show that

\( \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} g(x,y)f_{X,Y}(x,y)dxdy = E[g(X, Y)] = \int_{-\infty}^{+\infty} z f_{g(X,Y)}(z)dz \)
 
#16
Hello, BGM,

Yes, this time, your post exactly gave me what I was trying to look for early. Many thanks for your help. I have posted my computation when n=2 upstairs before I read your latest post. Thanks again for your great help.

DHB10
 
#17
Hello, BGM, Dason and Dragan and Other Madam/Sir,

How did you directly input all types of mathematical symbols such as integral symbol, sum symbol, etc.here in the Quick Reply message field? Is this way faster than that in MS Word ? because I have to use attachments to say what I want to say mathematically.

Another question: what does icon color beside a thread mean in this forum? some is red, some is green .

Thanks
DHB10
 
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#18
Hello, BGM,

For your latest post, regarding the dummy variables,

If f(z) = 2z + 3, then we have f(3t) = 2(3t) + 3 =6t+3. If f(x) = sin(x), then we get f(2u) = sin(2u)

If f(x,y) =2x+3y, then we have f(x, 2t) = 2x + 3(2t) =2x+6t. If f(x,y) = 4sin(x) + 5cos(y), then we obtain f(2u, 3u) = 4sin (2u) + 5cos (3u).

Because a function f(x) or f(x,y) symbol only shows the relationship between the resulting variable f(x) or f(x,y) and the variable(s). Are we talking the same thing regarding the dummy variables ?

Thanks

DHB10
 
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BGM

TS Contributor
#19
If you define \( g(z) = \int_{-\infty}^{+\infty} f(x, z) dx \), then \( x, z \) must be functionally independent.

The variable \( x \) is integrated out, and can be replaced by any arbitrary dummy variable

\( g(z) = \int_{-\infty}^{+\infty} f(x, z) dx = \int_{-\infty}^{+\infty} f(y, z) dy = \int_{-\infty}^{+\infty} f(w, z) dw = ... \)


You can use simple LaTeX code here wrapped by \( tag.\)
 
#20
Hello,

I am not sure if the integral limits are correct in my upstairs computation. If the integral limits is incorrect, the final result is incorrect for sure. I double checked them, I have not yet detected any mistake.

Thanks for yur help.

DHB10
 
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