# Suppose that $E[X^n] = 3n$. Find $E[e^X]$…

Suppose that $E[X^n] = 3n$. Find $E[e^X]$.

Hint from my professor: $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +···$

Not quite sure how to solve this problem, wouldn't $e^x$ go on exponentially. Any help is really appreciated.

• $\displaystyle\mathbb{E}[e^X] = \mathbb{E}[\sum_{k=0}^\infty \frac{X^k}{k!}] =\sum_{k=0}^\infty \mathbb{E}[ \frac{X^k}{k!}]$ where the last step needs to be... justified (at first do as if you could without any justification) – reuns May 10 '16 at 0:31
• Be careful that the formula does not hold for $n=0$, because $E[X^0]=E=1$ by definition. – user228113 May 10 '16 at 0:33

Warning: What appears below is at best vacuously true --- i.e. true but uninformatitve, since, as several people have pointed out, there is not actually any probability distribution whose $n$th moment, for every positive integer $n$, is $3n$.

\begin{align} \operatorname{E}\left(e^X\right) & = \operatorname{E}\left( 1 + X + \frac{X^2} 2 + \frac{X^3} 6 + \frac{X^4}{24} + \cdots \right) \\[10pt] & = 1 + \operatorname{E}(X) + \frac{\operatorname{E}(X^2)} 2 + \frac{\operatorname{E}(X^3)} 6 + \frac{\operatorname{E}(X^4)}{24} + \cdots \\[10pt] & = 1 + 3 + \frac 6 2 + \frac 9 6 + \frac{12}{24} + \cdots+ \frac{3n}{n!} + \cdots \\[10pt] & = 1 + \frac 3{0!} + \frac 3 {1!} + \frac 3 {2!} \cdots + \frac 3 {(n-1)!} + \cdots \\[10pt] & = 1 + 3\left( \frac 1 {0!} + \frac 1 {1!} + \frac 1 {2!} + \frac 1 {3!} + \cdots \right) \\[10pt] & = 1 + 3e. \end{align}

• This is formally correct - but only applies to nonexistent random variables. – Did May 14 '16 at 12:04

We show that there is no such random variable.

Note: we assume $E [X^n] = 3n$ for $n\ge 1$ (because, as commented above, the case $n=0$ cannot hold).

Recall that for real-valued random variable $Y$:

$$E( Y^2) \ge (E Y)^2$$

provided the second moment is finite (variance $E [Y^2- (E Y)^2] = E(Y^2)-(EY)^2$ is nonnegative). In particular, letting $Y=X^n$ for $n\ge 1$.

$$6n = E [X^{2n}] \ge (E [X^n])^2= (3n)^2,$$

or $n <6/9 <1$, so no such random variable exists !

• you are assuming that $E(Y)$ and $E(Y^2)$ are real ? or that $Y$ is a real r.v. ? – reuns May 10 '16 at 1:05
• +1. Already $E(X)=3$ and $E(X^2)=6$ lead to a contradiction. – Did May 10 '16 at 4:56
• Correct. Note that the result (nonexistence) holds even if you only assume the condition for n large enough. – Fnacool May 10 '16 at 13:10