What am i doing wrong in this Chi-square expectation problem?

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.

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

• Thanks for your answer. I just figured one mistake while I was solving the integral. Feb 2 at 11:38