I'm a little bit rusty on probability and expectation, and I'm practicing some exercises to take the rust off :)

In this case, let $X$ be a random variable with finite expectation $\mathbb{E}(X)$. Verify the following claims, motivating the answer:

  1. $\text{Var}\left[(X - \mathbb{E}(X)) \frac{1}{X}\right] = \frac{\mathbb{[E(X)]^2}}{\mathbb{E}[X^2] - [\mathbb{E}(X)]^2}$
  2. Let $\mathbb{E}[X]<0$ and $\theta\neq 0$ such that $\mathbb{E}(e^{\theta X}) = 1$; then $\theta>0$.

For the first point, one starts by using the properties of the variance on the l.h.s. and gets:

$\text{Var}\left( \frac{X - \mathbb{E}[X]}{X}\right) = \text{Var}\left( 1-\frac{\mathbb{E}[X]}{X}\right) = (\mathbb{E}[X])^2\text{Var}(\frac{1}{X}) = (\mathbb{E}[X])^2 \left( \mathbb{E}\left[\frac{1}{X^2}\right] - \left[ \mathbb{E} \left( \frac{1}{X} \right)\right]^2\right)$

which is clearly different from the r.h.s., since in general $\mathbb{E} [g(x)] \neq g(\mathbb{E}(X))$.

For the second point, my guess is that nothing can be said about $\theta$ since there is a contradiction; the moment generating function $\mathbb{E}(e^{\theta X})=1$ equals zero when differentiated, and we cannot subsequently plug in zero to obtain a negative value for the expectation, as the second claim affirms.

Are my observations correct? Am I missing something? Thanks to anyone willing to answer :)

Update: after the comment from idontgetit, I gave the second point some more tought, and came to the conclution that the claim must be false. We can consider the system of equations:

\begin{cases} x < 0 \\ e^{\theta x} = 1 \end{cases}

By noting that $x$ has to be less than zero, we can simplify everything by evaluating the inequality: $$ x < 1 - e^{\theta x} \rightarrow \frac{x}{1-e^{\theta x}}<0 $$ Since the above inequality is always true for the numerator, for the denominator we must have $$ 1 - e^{\theta x} > 0 \iff 1 > e^{\theta x} $$ which holds if, by using the logarithm properties, we observe that $$ 0 > \theta x $$ But since $x$ is always negative, $\theta$ can't possibly be greater than zero. Therefore, for the inequality to hold, $\theta<0$, and the claim in point 2 is also false. I hope to get some feedback even on this reasoning :)

  • 2
    $\begingroup$ To me, your reasoning in 1. looks correct. For 2, the statement does not tell you about the derivative of the MGF, only its value evaluated at a single point $\theta\neq 0$. Its derivative may well be non-zero. $\endgroup$
    – Idontgetit
    Commented Nov 23, 2022 at 13:55
  • 2
    $\begingroup$ for the second point, use Jensen's inequality: en.wikipedia.org/wiki/Jensen%27s_inequality (the map $x\mapsto e^{\theta x}$ is convex for all real $\theta$) $\endgroup$
    – Giulio R
    Commented Nov 23, 2022 at 15:52
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    $\begingroup$ Ok, using Jensen's inequality I get the same result, but is quicker. The system of inequalities that I set up came from exploiting the linearity property of the expectation. Thanks for the suggestion! $\endgroup$
    – james42
    Commented Nov 23, 2022 at 16:31


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