Maximum of Chi-square RV's

1. Jan 14, 2010

m26k9

Hello,
I am trying to find the distribution of the maximum of a set of independed Chi-square RV's with 2-degrees of freedom. Actually I only want to find the MEAN value.
I am using the following formula to find the PDF.

$$f_{X_{\mathsf{max}}}(x) = NF_X(x)^{N-1}f_X(x)$$

Following PDF and CDF is used:
$$f_X(x)=\frac{1}{2}e^{-\frac{x}{2}}$$
$$F_X(x) = 1-e^{-\frac{x}{2}}$$

So what I want to find is: (Assuming N variables)
$$E[f_{X_{MAX}}(x)] =N \int_0^R xF_X(x)^{N-1}f_X(x)$$

I am stuck in a neverending integration by-parts.
If anybody know any solution to this or any method to find this, please let me know.

Cheer.s

Last edited: Jan 14, 2010
2. Jan 15, 2010

bpet

Did you mean to say $$E[X_{MAX}]$$.

Mathematica gives the answer 2*HarmonicNumber[n] but I'm not sure how to derive it.

3. Jan 15, 2010

m26k9

Thank you bpet.
Yes I want to find $$E[X_{MAX}]$$. Sorry for the mistake.
I will search for Harmonicanumber. Havent heard of it before.

Cheers.

4. Jan 24, 2010

EngWiPy

You need to calculate the mean of the maximum of independent Chi-square RVs, is it right? So:

$$E_{X^*}[X]=\,\int_0^{\infty}x\,f_{X^*}(x)\,dx$$

where $$f_{X^*}(x)$$ is the first equation you wrote, but the PDF and CDF must be of Chi-square not of exponentials. Then substiute these data into the integration and evaluate the integral.

5. Jan 24, 2010

m26k9

Thank you very much David.
I'm not sure what you meant by but the PDF and CDF must be of Chi-square not of exponentials?

Because of the exponentials I could not find a closed-from expression there is a pattern and I could find a recursive solution. Could you please explain a bit what you meant earlier?

Cheers.

6. Jan 24, 2010

EngWiPy

I mean the PDF and CDF in the first equation must be of a Chi-square random variable. To elaborate, suppose that we have n independent and identically distributed Chi-square random variables: $$X_1,\,X_2,\ldots,\,X_n$$ with $$f_X(x)$$ and $$F_X(x)$$ as the PDF and CDF, respectively. Arrange them in ascending order as: $$X_{(1)}\leq X_{(2)}\leq X_{(3)} \leq \cdots\leq X_{(n)}$$. Then the PDF of $$X_{(n)}$$ (the maximum RV) is:

$$f_{X_{(n)}}(x)=\,n\left[F_X(x)]^{n-1}\,f_X(x)$$

You can find the distributions of a Chi-square RV from any probability book.

7. Jan 24, 2010

bpet

m26k9's expression for the distribution is correct.

8. Jan 24, 2010

m26k9

Thank you very much David and Bpet.

The problem I had was to evaluate the integral and was thinking if there is a commonly known closed-form expression for $$f_{X_{(n)}}(x)$$. I guess I have to do with the recursive one.

Thank you again guys.

9. Jan 25, 2010

EngWiPy

Yes, bpet is right, I am sorry, where Chi-square distributions with two degrees of freedom become as m26k9 wrote with the assumption that the variance is unity. So, the PDF of the maximum RV becomes:

$$f_{X_{(n)}}(x)=\,n\left[F_X(x)\right]^{n-1}\,f_X(x)=\,\frac{n}{2}\left[1-\text{e}^{-x/2}\right]^{n-1}\,\text{e}^{-x/2}$$.

Where using binomial expansion:

$$\left[1-\text{e}^{-x/2}\right]^{n-1}=\sum_{k=0}^{n-1}(-1)^k{n-1\choose k}\text{e}^{-kx/2}$$

After simple manipulation, use the table of integral to solve the resulting integral.

Good luck