Truthfulness of statements on the expected values of random variables Are these statements true or false? Why?


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*$E(|X|)\le 1 + E(X^2)$


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


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*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.
 A: 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 $x \to 1 + \theta x$.  

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.
A: 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?
