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Let $X_{1},X_{2},..,X_{10}$ be 10 standard normal variates. Let us Define $T = X_{1}^{2}+X_{2}^{2}+..+X_{10}^{2}$. It is required to compute $E(\frac{1}{T})$.

Now, We know that T (sum of squares of standard normal) will follow chi-square with 10 dof. So, we can compute the expectation as:

$E(\frac{1}{T}) = \int_{0}^{\inf}\frac{1}{gamma(5)2^{5}}x^{5-1}e^{-\frac{x}{2}} * \frac{1}{x}dx$

Using the gamma integram, the above equation simply gives me $\frac{1}{2}$. But i am not getting the same answer as the answer key.

Someone please review my steps.

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$$E[1/T]=\frac{1}{\Gamma(5)2^4}\int_0^\infty {1\over 2}x^3 e^{-x/2}dx$$ The integral is the third moment of exponential random variable with $\lambda=1/2$, and it is $3!/\lambda^3$. Therefore, the result becomes: $$E[1/T]=\frac{3! 2^3}{\Gamma(5)2^4}=\frac{1}{8}$$

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  • $\begingroup$ Thanks for your answer. I just figured one mistake while I was solving the integral. $\endgroup$
    – userNoOne
    Feb 2 at 11:38

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