# Show that the variance of this random variable is finite

Let $X$ be a random variable such that $P(X<-1)=0$. Assume also that $E[X]$ and $Var[X]$ are finite.

Show that $Var[\log(1+\frac{1}{2}X)]$ is finite.

I haven't been able to show this. Maybe this is not possible?

Update: It seems to me that I don't even need the $Var[X]$ to be finite. If I can show that $(\log(1+x))^2 < x$ for all $x > 0$ then it seems I'm basically done, since $Var[X]=E[X^2] - (E[X])^2$. The problem is showing that inequality. What do you think?

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@JohnEngbers Well, it seems to me that I don't even need the $Var[X]$ to be finite. If I can show that $(\log(1+x))^2 < x$ for all $x > 0$ then it seems I'm basically done, since Var[X] is nothing but $E[X^2] - (E[X])^2$. The problem is showing that inequality. What do you think? – meh Jun 25 '12 at 17:27
Seems promising. But I'd write $\log(1+x) < x$ for $x>0$, so $\log(1+x/2) < x/2$ and $(\log(1+x/2))^2 < x^2/4$ and from that we can bound both the mean and variance (but we need to know that Var(X) is finite). – leonbloy Jun 25 '12 at 17:46

Since $x > 0$, your inequality can be re-stated as $f(x) = 1+x \leq e^{\sqrt{x}} = g(x)$. Observe that $f(0) = g(0) = 1$. Hence, it is enough to show that $f'(x) < g'(x)$ for all $x > 0$. That is,

$$1 \leq \frac{1}{2\sqrt{x}}e^{\sqrt{x}}$$

Calling $\sqrt{x} = y$,

$$a(y) = 2y \leq e^{y} = b(y)$$

Now, for $y \leq 1$, observe that $e^{y} \geq 1+y \geq 2y$. For $y > 1$, it is enough to show that $a'(y) \leq b'(y)$, that is $2 \leq e^{y}$. Since $y > 1$, $e^{y} > e > 2$, which completes the proof.

Back to the main question, recall that:

$$E[\log(1+0.5X)^{2}] = E[I(X \geq 0)\log(1+0.5X)^{2}] + E[I(-1 \leq X \leq 0)\log(1+0.5X)^{2}]$$

Hence, it is enough to show that both terms on the RHS are finite. Using your inequality, the following holds with probability $1$,

$$0 \leq I(X \geq 0)\log(1+0.5X)^{2} \leq I(X \geq 0)0.5X$$

and

$$E[I(X \geq 0)\log(1+0.5X)^{2}] \leq E[I(X \geq 0)0.5X] < \infty$$

Since $I(-1 \leq X \leq 0)\log(1+0.5X)^{2}$ is bounded, the proof is complete.

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