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

BGM

TS Contributor
#21
A mistake in the integration order: When the inner integral limit is \( \int_x^{+\infty} \), it must be with respect to \( dy \);
If you integrate with respect to \( dx \) first, the corresponding inner integral limit should be \( \int_{-\infty}^y \).

i.e. \( \int_{-\infty}^{+\infty} \int_x^{+\infty} f(x, y) dydx = \int_{-\infty}^{+\infty} \int_{-\infty}^y f(x, y) dxdy \)

Changing the integration order will also change the limits.

Also I am not sure if you can directly claim \( I_{1,1} = 0 \)
 
#22
Hello, BGM,

I am going to try again as what you instructed. Because I(1,1) is the same as E(XY) when n=2, So, I directly claim I(1,1) = 0 ( it is my thinking)

Thanks a lot
DHB10
 
#25
Hello,

Based on the correct answer to E(R^2) when n =3 ( see the table in my first post), there must be a sqrt(3) created in the final result of E(XY), it seems that there is nowhere to create a sqrt(3) for the result of E(XY). So, I am wondering if the whole approach is correct.

Thanks
DHB10
 

BGM

TS Contributor
#26
When you are computing \( I_{1,2} \), the integration domain should be \( - \infty < x < y < t < +\infty \)

So when you try to change to the polar coordinate, the inner double integral should become the following:

When \( x < 0 \),
\( \int_x^{+\infty} \int_y^{+\infty} f(y, t) dtdy \) \(= \int_{\frac {\pi} {4}}^{\frac {\pi} {2}}\int_0^{+\infty} f(r\cos\theta, r\sin\theta) rdrd\theta + \int_{\frac {\pi} {2}}^{\frac {5\pi} {4}} \int_0^{\frac {x} {\cos\theta}}f(r\cos\theta, r\sin\theta) rdrd\theta\)

When \( x > 0 \),
\( \int_x^{+\infty} \int_y^{+\infty} f(y, t) dtdy \) \(= \int_{\frac {\pi} {4}}^{\frac {\pi} {2}}\int_{\frac {x} {\cos\theta}}^{+\infty} f(r\cos\theta, r\sin\theta) rdrd\theta\)

Also whenever you get \( \int_{-\infty}^{+\infty} xf_X(x)dx = E[X] \), it should be correspond to the original, unordered sample, so in your case it should be 0.
 
#27
Hello, BGM,

Yes, your conclusion in the post upstairs is correct, because I checked my integration again, there was a minus sign (-) missing on page 3 of 4 in my previous post. So, the final result E(XY) = 0 , when n=3 according to this approach. Please see the attachment below ( the other pages are totally the same as previously posted) So, I only post the last part this time).

So, if E(XY) = 0, when n=3 , in this case, then the computed E(R^2) is incorrect compared with the one in the Table 2 in my first post for this topic. In fact, the result in the said table is correct except the one E(R^2) when n=4. you can simulate it to verify it.

So, what is the whole logical mistake located for this approach ?

Thanks a lot
DHB10
 
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#28
Hello,

I have updated the page 3 and 4 in previous post upstairs to add the missing minus sign(-). The outcome is now E(XY) = 0 when n =3 based on this approach accordingly. However, the correct answer should not be E(XY) =0 in this case. So, where is the generally logical mistake located ? or where is the conceptually logical mistake located in this case?

Thanks
DHB10
 
#30
Hello,

The unknown author said in the page upstairs that: "The moments of the range R can be derived from either the pdf above or from the moments of the minimum and maximum order statistics X(1) and X(n)....."

If we derive E(R^2) directly from the given pdf of the R in the page shown upstairs, the evaulation of the integrals would be extremely very hard . I believe that the author used the moments of X(1) and X(n) to indirectly compute the second moment of R , E(R^2).

This is why I am trying to find out a way to compute the second moment of R, E(R^2). Unfortunately, I get depressed my several trials have been a failure even my final computation is correct using the approach being disscussed , which I guss that there must be a conceptually logical mistake somewhere. To find out a logical mistake is not easy especially for the one who created the logical mistake initially. Please help with this issue.

Thanks a lot
DHB10
 
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#31
Hello, BGM,

For your post dated on Yesterday 11:07 AM (#35), in my opinion, we don't need to split the integration when computing I(1,2). Please see the following derivation for the reason.

Thanks
DHB10
 
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BGM

TS Contributor
#32
Of course I know why \( \frac {\pi} {4} < \theta < \frac {5\pi} {4} \)

It is correspond to the inequality \( t > y \) only

So you missed the inequality \( y > x \) here.

Please see the depicted image for the reasoning. It is hard for me to describe the picture by word.

如有需要用中文討論也可 :)
 
#33
Hello, BGM,

Let me study your graphs and then I am going to reply to your latest post. Many thanks for your patience & help to take care of this thread.

Thanks
DHB10
 
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#34
Hello, BGM,

Another question, up to now, according to both your conclusion based on the probablity theory and my final computation with your help, when n = 3, E(XY) = 0 in this case. However, in this case, E(XY) should not be equal to zero , but be sqrt(3)/PI. Where is the conceptually logical mistake located in this case ?

Based on my several times computation, by my common sense, this is no place to creat any sqrt(3) from the beginning to the end using this approach.

Thanks a lot.
DHB10
 
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BGM

TS Contributor
#35
Now given that \( E[X_{(1)}^2] = E[X_{(n)}^2] = 1 + \frac {\sqrt{3}} {2\pi} \), \( E[R^2] = 2 + \frac {3\sqrt{3}} {\pi} \) (verified by simulation)

So we expect

\( E[R^2] = E[(X_{(n)} - X_{(1)})^2]
= E[X_{(n)}^2 + X_{(1)}^2 - 2X_{(n)}X_{(1)}] \) \(
= E[X_{(n)}^2] + E[X_{(n)}^2] -2E[X_{(n)}X_{(1)}]\)

\( \Rightarrow E[X_{(n)}X_{(1)}]
= \frac {1} {2} (E[X_{(n)}^2] + E[X_{(1)}^2] - E[R^2])
= \frac {1} {2} \left(1 + \frac {\sqrt{3}} {2\pi} + 1 + \frac {\sqrt{3}} {2\pi} - 2 -\frac {3\sqrt{3}} {\pi}\right) \) \( = -\frac {\sqrt{3}} {\pi} \)

This result should be ok as we expect that they are negatively correlated.

So you can check against this number to see if you can get \( -\frac {\sqrt{3}} {\pi} \)

From the previous discussion, at least I think your integration limits is incorrect after changing to polar coordinates. So I have not conclude it is 0 yet.
 
#36
Hello, BGM,

Yes, E(R^2) = minus sqrt(3)/PI is the correct result. but it seems that there is no place to create a sqrt(3) in the computation. I am going to check my integration again to see if there is any calculus computation mistake, focusing on the the integration limitation check, especially when changing to polar coordinates.

Thanks a lot.
DHB10 from China, PRC
 

BGM

TS Contributor
#37
I think you have get my point as you can draw the picture by yourself.

Maybe I further elaborate my idea:

Consider the \( y-t \) plane:

When \( x < 0 \), and when \( \frac {\pi} {2} < \theta < \frac {5\pi} {4} \), we require that \( x < y < 0 \). By changing to polar coordinates
\( y = r\cos\theta \), we have \( x < r\cos\theta < 0 \Rightarrow 0 < r < \frac {x} {\cos\theta} \) as \( \cos\theta < 0 \) when \( \frac {\pi} {2} < \theta < \frac {5\pi} {4} \)

As what you have said above, it seems that it cannot help to evaluate the triple integral. Maybe the spherical coordinates can help, but I am not sure. I have not try yet.
 
#38
Hello, BGM,

Thanks for your help and your elaborate idea upstairs. I am going to try it. Because I don't think that there is a conceptuallly logical mistake for this approach to compute the E(XY). Even if there is some pure computation mistakes, we can detect it after double checking.

Thanks a lot.
DHB10
 

BGM

TS Contributor
#39
I think integrating with respect to \( t \) first is difficult because you need to deal with the integral like \( t^n e^{-\frac {1} {2}t^2} \) where \( n \) is even - which involves the CDF of normal and has no closed form solution for you to continue to integrate. However when \( n \) is odd you just reduced to a gamma integral. So you may try to change the integration order:

Recall \( -\infty < x < t < y < +\infty \)

\( E[XY] = \frac {6} {(2\pi)^{\frac {3} {2}}}
\int_{-\infty}^{+\infty} \int_x^{+\infty} \int_x^y xye^{-\frac {1} {2}(x^2+y^2+t^2)}dtdydx \)

\( = \frac {6} {(2\pi)^{\frac {3} {2}}} \int_{-\infty}^{+\infty}
e^{-\frac {t^2}{2}}\int_t^{+\infty} ye^{-\frac {y^2} {2}}\int_{-\infty}^t
xe^{-\frac {1} {2} x^2}dxdydt \)

\( = -\frac {6} {(2\pi)^{\frac {3} {2}}} \int_{-\infty}^{+\infty}
e^{-\frac {t^2}{2}-\frac {t^2}{2}} \int_t^{+\infty} ye^{-\frac {y^2} {2}} dydt \)

\( = -\frac {6} {(2\pi)^{\frac {3} {2}}} \int_{-\infty}^{+\infty}
e^{-\frac {3t^2}{2}} dt \)

\( = -\frac {6} {(2\pi)^{\frac {3} {2}}} \sqrt{\frac {2\pi} {3}} \)

\( = - \frac {\sqrt{3}} {\pi} \)

Previously I am too busy and overlook these stuffs, so actually for \( n = 3 \) case it is pretty straight forward.
 

BGM

TS Contributor
#40
Following the similar idea, for \( n = 4 \), let \( X < T < W < Y \) be the order statistics and \( -\infty < x < t < w < y < +\infty \) be the support.

The joint pdf is \( \frac {4!} {4\pi^2} e^{-\frac {1} {2} (x^2 + y^2 +t^2 +w^2)} \)

So \( E[XY] = \frac {6} {\pi^2} \int_{-\infty}^{+\infty} \int_t^{+\infty} e^{-\frac {1} {2} (t^2 + w^2)} \int_w^{+\infty} ye^{-\frac {y^2} {2}} \int_{-\infty}^t xe^{-\frac {x^2} {2}} dxdydwdt \)

\( = - \frac {6} {\pi^2} \int_{-\infty}^{+\infty} \int_t^{+\infty} e^{-(t^2 + w^2)} dwdt \)

\( = - \frac {6} {\pi^2} \times \frac {2\pi} {2} \times \frac {1} {2} \)

\( = - \frac {3} {\pi} \)

And this is verified by simulation.

Combining together we have \( E[R^2] = 2 + \frac {6 + 2\sqrt{3}} {\pi} \) for \( n = 4 \)