Let $X$ be a random variable with expected value $\mathbb{E}(X)<+\infty$.
Prove if the following equivalence relations are true or not:
- $\mathbb{E}(|X|)\leq 1 + \mathbb{E}(X^2)$
- $\text{Var}\left[\frac{X-\mathbb{E}(X)}{\mathbb{E}(X)}\right] = \frac{\mathbb{E}(X^2)-[\mathbb{E}(X)]^2}{[\mathbb{E}(X)]^2}$
- $\text{Var}\left[\frac{X-\mathbb{E}(X)}{X}\right] = \frac{[\mathbb{E}(X)]^2}{\mathbb{E}(X^2)-[\mathbb{E}(X)]^2}$
- If $\mathbb{E}(X)<0$ and $\vartheta\neq0$, so that $\mathbb{E}(e^{\vartheta X}) = 1$, it must be $\vartheta>0$
I'm having problems with the fourth but I guess I had the right intuition. Anyway, here's how I solved them. In addition to help me with the last one, I'd be grateful if you could check if I'm right on the others and/or suggest more straightforward (or elegant!) way to proceed.
- TRUE
Since $\mathbb{E}(X^2) = \mathbb{E}(|X|^2)$ and $1=\mathbb{E}(1)$, to prove $\mathbb{E}(|X|)\leq 1 + \mathbb{E}(X^2)$ is the same as to prove $$\mathbb{E}(1+|X|^2 - |X|)\geq0$$ Since $1+|X|^2 - |X|\geq0,\,\forall\,X$, also its expected value has to be always non-negative.
- TRUE
Since $\frac{1}{\mathbb{E}(X)}$ is a scalar, it can be applied the well-known relation $\text{Var}(aX) = a^2 \text{Var}(X)$ (where $a$ is indeed a scalar), moreover we'll use the obvious relation $\text{Var}(k)=0$ if $k$ is a constant, so to find $$\text{Var}\left[\frac{X-\mathbb{E}(X)}{\mathbb{E}(X)}\right] = \frac{1}{\left[\mathbb{E}(X)\right]^2}\text{Var}\left[X-\mathbb{E}(X)\right] = \frac{\text{Var}\left[X\right]}{\left[\mathbb{E}(X)\right]^2} = \frac{\mathbb{E}(X^2)-[\mathbb{E}(X)]^2}{[\mathbb{E}(X)]^2}$$
- FALSE
Since the variance of the difference of two random variables is the sum of the variances, we find $$\text{Var}\left[\frac{X-\mathbb{E}(X)}{X}\right] = \text{Var}\left[1-\frac{\mathbb{E}(X)}{X}\right] = \text{Var}\left[\frac{\mathbb{E}(X)}{X}\right] = \left[\mathbb{E}(X)\right]^2 \text{Var}\left[\frac{1}{X}\right] \neq \left[\mathbb{E}(X)\right]^2 \frac{1}{\text{Var}(X)}$$
- ...?
If we write $$e^{\vartheta X} = \sum_{k=0}^\infty \frac{(\vartheta X)^k}{k!} = 1 + \left( \vartheta X \right) + \frac{\left( \vartheta X \right)^2}{2} + \frac{\left( \vartheta X \right)^3}{3!} + \dots$$ it follows immediately that $$\mathbb{E}(e^{\vartheta X}) = 1 + \mathbb{E}\left(\vartheta X \right) + \mathbb{E}\left(\frac{\left( \vartheta X \right)^2}{2}\right) + \mathbb{E}\left(\frac{\left( \vartheta X \right)^3}{3!}\right) + \dots$$ and since we know that $\mathbb{E}(e^{\vartheta X}) = 1$ all the terms of order greater than zero must cancel out. In order for this to be true one necessary condition is that they don't hold the same sign...
You can use Jensen's inequality. Note that both $e^x$ and $e^{-x}$ are both convex functions in $x$. Hence for $\theta \ne 0$, $$E[e^{\theta X}] \ge e^{\theta E[X]}$$
Since $E[X]$ is negative, $\theta <0$ will give rhs strictly greater than 1. Therefore the opposite must hold.