# If $X$ has a Poisson distribution with $E[X]=\lambda$, does $Var[X^2]=4\lambda^3+6\lambda^2+\lambda$?

Suppose $X$ has a Poisson distribution with mean (and therefore variance) $\lambda$. Using Excel to explore properties of the distribution of $X^2$ with some small integer values of $\lambda$ I found:

The values of $E[X]$ are consistent with the formula (which was given in an answer to this question):

$$E[X^2] = \lambda^2 + \lambda$$

Trying to find a formula to fit the values of $Var[X^2]$, I came up with:

$$Var[X^2] = 4\lambda^3+6\lambda^2+\lambda$$

Is this formula generally valid, and if so how can it be proved?

We can use the fact that

$$Var(X^2)= \mathbb{E}[X^4]-\mathbb{E}[X^2]^2$$

You can explicit $\mathbb{E}[X^4]$ either through the definition or using the moment-generating function of a Poisson-distributed random variable. I will use the latter. The moment-generating function $M_X(t)$ of a Poisson-distributed random variable $X$ is

$$M_X(t)=e^{\lambda(e^t-1)}$$

Now, $$\mathbb{E}[X^4]=\frac{d^4}{dt^4}M_X(0)=|\lambda e^{\lambda(e^t-1)+t}(\lambda^3 e^{3t}+6\lambda^2 e^{2t}+7\lambda e^t+1)|_{t=0}=\lambda^4+6\lambda^3+7\lambda^2+\lambda$$

$$\mathbb{E}[X^2]=\frac{d^2}{dt^2}M_X(0)=|\lambda e^{\lambda(e^t-1)+t}(\lambda e^t+1)|_{t=0}=\lambda^2+\lambda \implies \mathbb{E}[X^2]^2=\lambda^4+ 2\lambda^3+\lambda^2$$

We can now determine the variance of $X^2$.

$$Var(X^2)=\mathbb{E}[X^4]-\mathbb{E}[X^2]^2=\lambda^4+6\lambda^3+7\lambda^2+\lambda-\lambda^4-2\lambda^3-\lambda^2=\color{red}{4\lambda^3+6\lambda^2+\lambda}$$

• Why did you downvote my answer? – Lonidard May 22 '15 at 13:49
• Isn't it basically a rephrasing of the two other answers? – Clement C. May 22 '15 at 13:51
• It simply took me a couple of minutes more to write the answer because I wanted to get more in depth than you did. Since this is a quite basic question, I thought showing some more work could have helped the OP. I believe Stackexchange is a place where people help each other, not a rep rush for who answers the question first. – Lonidard May 22 '15 at 14:01
• It is helpful, thank you ... but it's also fair to note that the more concise answers by Clement C. and Jack D'Aurizio were posted first. – Adam Bailey May 22 '15 at 14:56
• Yes, indeed. I don't see a correlation with downvoting, though. – Lonidard May 22 '15 at 15:17

Hint: you can rewrite $$\operatorname{Var}[X^2] = \mathbb{E}[X^4] - \mathbb{E}[X^2]^2$$ and use the known expressions of the (raw) moments of a Poisson-distributed random variable (or recompute them — you already can get $\mathbb{E}[X^2]$ from the variance and the expectation).

The moment generating function of a Poisson random variable $X$ with mean $\lambda$ is given by: $$f(t) = \mathbb{E}[e^{tX}] = e^{\lambda(e^t-1)},\tag{1}$$ hence:

$$\begin{eqnarray*}\text{Var}[X^2]=\mathbb{E}[X^4]-\mathbb{E}[X^2]^2 &=& f^{(4)}(0)-f''(0)^2\\&=&(\lambda +7 \lambda ^2+6 \lambda ^3+\lambda ^4)-(\lambda+\lambda^2)^2\tag{2}\end{eqnarray*}$$ proving your claim.