Are these statements true or false? Why?

  • $E(|X|)\le 1 + E(X^2)$

$0≤|x|<1+x^2$ for all choices of $x$ with $x$ real number. What with $X$ random variable?

  • if $E(X)<0$ and $ \theta \neq0$ such that $E[e^{\theta X}]=1 $, then $\theta>0$

I try to evaluate whether the assertion is true or false:

$E[e^{\theta X}]=1$ then

$e^{\theta E[ X]}=1 $

$\theta E[ X] = 0$

Since $ \theta \neq0, E[X] $ must be equal to zero and cannot be less than 0 as the exercise says.

Edit: I noted that the equality $E[e^{\theta x}]=e^{θE[X]}$ doesn't hold in general, and I have no idea how to solve this question/find a counterexample to prove that it is false.

  • 1
    $\begingroup$ Are these questions from a course or textbook? Please add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. Please make these changes as just posting your homework & hoping someone will do it for you is grounds for closing. If they aren't questions from a course or textbook, it would be interesting to know what motivated you to ask - is it something you came across in a proof for instance? $\endgroup$
    – Silverfish
    Jul 13, 2016 at 12:41
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    $\begingroup$ Forget random variables for a while. For real number $x$, is it true that $|x| \leq 1 + x^2$ for all choices of $x$? $\endgroup$ Jul 13, 2016 at 13:13
  • $\begingroup$ Yes, $|x|≤1+x^2$ for all choices of $x$. $\endgroup$
    – Motmot
    Jul 13, 2016 at 13:27
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    $\begingroup$ Thanks for making those changes to show your attempts at the question and for adding the self-study tag. $\endgroup$
    – Silverfish
    Jul 13, 2016 at 13:30
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    $\begingroup$ It is not the case that $E[e^{\theta X}] = e^{\theta E[X]}$. In fact, that is true only when $X$ is almost surely constant or $\theta=0$. $\endgroup$
    – whuber
    Jul 13, 2016 at 17:08

2 Answers 2


Because $\exp$ is convex at $0$, the graph of $x \to e^{\theta x}$ lies above its tangent line at $0$ (strictly above for $\theta\ne 0$), which has formula $x \to 1 + \theta x$.

Figure showing graphs of the two functions

The solid blue curve graphs $x\to e^{-x/2}$, depicting the case $\theta=-1/2$. The dotted red line is the tangent to the blue curve at $x=0$. Its equation is $1 + \theta x = 1 - x/2$.

This proves that

$$e^{\theta x} \ge 1 + \theta x$$

for all $\theta$ and all $x$.

Assuming $\theta$ is a constant for which $\mathbb{E}(e^{\theta X})=1$, we may use this observation to obtain a lower bound

$$0 = -1 + \mathbb{E}(e^{\theta X}) \ge -1+\mathbb{E}(1 + \theta X) =\theta\,\mathbb{E}(X).$$

Therefore $\theta$ and $\mathbb{E}(X)$ must have opposite signs. In particular, if $\mathbb{E}(X) \lt 0$, then $\theta \ge 0$. Since $\theta=0$ has been explicitly ruled out, we conclude $\theta \gt 0$, QED.


Since $0\leq|x|<1+x^2$ we have that $$E|X|=\int |x|dF(x)<\int(1+x^2)dF(x)=E(1+X^2)$$ where $F$ represents the density function.

I am not sure how to understand your second question, in particular I do not see the "IF..., THEN ..." statement :-) Would you mind editing it a bit?

  • $\begingroup$ Thank you for the answer to the first one. I edited the second one. $\endgroup$
    – Motmot
    Jul 13, 2016 at 15:05

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