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Problem in my Stat class

Let $Y_1, Y_2,\ldots, Y_n$ be arandom sample of size $n$ from

$$f_Y(y)= \frac{1}{\theta } e^{-y/\theta}, \qquad y> 0,\theta > 0$$

Use moment-generating functions to show that the ratio $2n\bar{y}/\theta$ has a chi square distribution with $2n$ degrees of freedom.

First question do they mean that $2n\bar{y}$ is the $y$ variable and $\theta$ is the $\theta$ variable?

secondly how do you show that anything has a chi squared distribution?

Small hints to guide me in the right direction would also be helpful.

Michael Hardy
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Postskjerm
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    (Big) hint: what is $\bar y$? – Did Sep 24 '16 at 16:11
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    Now I understand that I've misunderstood the question, they want me to show that the ratio has chi distribution not $f(y)$ – Postskjerm Sep 24 '16 at 16:19
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    There is a reason why in the expression $f_Y(y)$, one $Y$ is capital and the other is in lower case. I edited acccordingly. Note that the expression $\Pr(Y\le y)$ could otherwise not be understood. $\qquad$ – Michael Hardy Sep 25 '16 at 17:16

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Note that your given distribution for $Y_1,\dots,Y_n$ is that of an exponential family. In particular, $$ M_Y(t) = E[e^{tY}] = \int_{0}^\infty e^{ty} \left( \dfrac{1}{\theta} e^{-y / \theta} \right) \; dy = \cdots = \dfrac{1}{1-\theta t} \text{ for } \dfrac{1}{\theta} > t. $$ Let $Z = \dfrac{2 n \overline{Y}}{\theta} = \dfrac{2}{\theta} \displaystyle\sum_{k=1}^n Y_i$. It follows that \begin{align*} M_Z(t) = E[e^{tZ}] &= E \left[ \exp \left( \dfrac{2t}{\theta} \displaystyle\sum_{k=1}^n Y_i \right) \right] \\ &= E \left[ \displaystyle\prod_{k=1}^n \exp \left( \dfrac{2t}{\theta} Y_i \right) \right] \\ &= \displaystyle\prod_{k=1}^n E \left[ \exp \left( \dfrac{2t}{\theta} Y_i \right) \right] \quad \text{(Independence)} \\ &= \left( E \left[ e^{2tY/\theta} \right] \right)^n \quad \text{(Identically Distributed)} \\ &= (M_Y(2t/\theta))^n \\ &= \left( \dfrac{1}{1-\theta (2t/\theta)} \right)^n, \quad \text{ with } \frac{1}{\theta}>\left( \dfrac{2t}{\theta} \right) \iff t < \frac{1}{2} \\ &= \left( \dfrac{1}{1-2t} \right)^n, \quad t < \frac{1}{2} \end{align*} which is precisely the mgf of a chi-square distribution with $2n$ degrees of freedom.

user365239
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