# Claims on expectation and MGF

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 :)

• 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. Commented Nov 23, 2022 at 13:55
• 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 $\theta$) Commented Nov 23, 2022 at 15:52
• 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! Commented Nov 23, 2022 at 16:31