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Let $Y$ be a random variable with an exponential distribution with E(Y)=$\theta$.

Show that the pivotal $\frac{2Y}{\theta}$ has a chi-square distribution using moment-generating function technique.

What I did: Mu(t) = E(e$^{tu})$ = E(e$\frac{t(2Y)}{\theta})$.

Then I integrated: $\int_{0}^{\infty}$e$^\frac{t(2Y)}{\theta}*f(y) dy$.

This produces $\frac{-\theta}{2 \beta t-\theta}$

This looks nothing like a chi-square distribution. It then asks how many degrees of freedom it has. Please help! Thanks in advance, oh wise Stats geniuses.

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Hints:

  • What is $\beta$?

  • I suspect when you wrote $\dfrac{-\theta}{2 \beta t-\theta}$ you may have intended $\dfrac{-\theta}{2 \theta t-\theta}$, which can be simplified

  • A chi-squared distribution with $k$ degrees of freedom has a moment generating function of $$(1-2t)^{-k/2}$$

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Where did your $\beta$ come from in your result?

$$ \int_0^{\infty}\frac{e^{\frac{t(2y)}{\theta}}e^{-\frac{y}{\theta}}}{\theta}dy=\frac{1}{1-2t} $$

Perhaps you made a small mistake in your calculus.

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Letting $P \equiv 2Y / \theta$ denote your pivotal quantity, you have:

$$\begin{equation} \begin{aligned} m_P(t) \equiv \mathbb{E}(\exp(tP)) &= \mathbb{E} \Big( \exp \Big( \frac{2tY}{\theta} \Big) \Big) \\[6pt] &= \int \limits_0^\infty \frac{1}{\theta} \cdot \exp \Big( \frac{2ty}{\theta} \Big) \cdot \exp \Big( - \frac{y}{\theta} \Big) dy \\[6pt] &= \frac{1}{\theta} \int \limits_0^\infty \exp \Big( - \frac{(1-2t)}{\theta} \cdot y \Big) dy \\[6pt] &= \frac{1}{1-2t} \Bigg[ \exp \Big( - \frac{(1-2t)}{\theta} \cdot y \Big) \Bigg]_{y=0}^{y \rightarrow \infty} \\[6pt] &= \begin{cases} (1-2t)^{-1} & \text{ for } t < 1/2, \\[6pt] \infty & \text{ for } t \geqslant 1/2. \end{cases} \end{aligned} \end{equation}$$

This result is the moment-generating function for a Chi-squared random variable with two degrees-of-freedom, so we have the distribution $P \sim \chi_2^2 $.

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