$\chi^2$ function problem - moment generating functions

Question

Consider $X$ to be such that $X$ is distributed $\chi_m^2$ for a natural number $m$. I want to do the following problems:

(i) Prove that the MGF (Moment Generating Function) of $X$ is $m_X(t)=\frac{1}{(1-2t)^{\frac{m}{2}}}$.

(ii) Considering the sequence of independent $\chi_1^2$ variables $A_1,\cdots $, where $\bar{A_n}=\frac{A_1+\cdots A_n}{n}$, to prove $\frac{\bar{V_n}-1}{\sqrt{\frac{2}{n}}}\rightarrow Z$ where $Z$ is a standard normal variable.

Progress: The chi distribution is continuous. I have tried using the definition of the moment generating function, that $E(X^k)=\int_{-\infty}^{\infty}x^kf_X(x)dx$. I did not obtain an equivalent expression to the one given though. For (ii) I suspected the central limit theorem would be of use but this did not work out - any solutions would be awesome:) just trying to figure out how to do these sorts of problems to be honest.

Answer 1

• The definition of $m_X$ is $m_X(t)=E[e^{tX}]$.
• If $Z_1, \ldots, Z_m$ are i.i.d. standard normal, then $X := Z_1^2 + \cdots + Z_m^2$ follows the $\chi^2_m$ distribution.
• Justify why $E[e^{tX}] = E[e^{tZ_1^2}] \cdot E[e^{tZ_2^2}] \cdots E[e^{tZ_m^2}] = (E[e^{tZ_1^2}])^m$ holds.
• Show that for $t < 1/2$ we have $E[e^{tZ_1^2}] = \frac{1}{\sqrt{1-2t}}$ by integrating $\int_{-\infty}^\infty e^{tz^2} \frac{1}{\sqrt{2\pi}} e^{-z^2/2} \, dz$. (Hint: try the change of variables $u=z\sqrt{1-2t}$.)

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Part (ii) is a direct application of the central limit theorem.