I would really appreciate anyone's help with this problem:

(let $E$ denote expectation)

Suppose $X$ and $Y$ are independent Poisson random variables, each with mean $1$.
Find: $E[(X + Y)^2]$ "

My question is why can't I expand the $(X + Y)^2$ to get $E(X^2 + 2XY + Y^2)$ and then use linearity to get $E(X^2) + 2E(X)E(Y) + E(Y^2)$. Then since $X$ and $Y$ are independent, this would give $E(X)^2 + 2(1)(1) + E(Y)^2 = 1 + 2 + 1 = 4$. But the answer is 6.

I can get the correct answer through this method: $E[(X+Y)^2] = Var(X + Y) + E[(X+Y)]^2$ and noting that $X + Y$ has a Poisson distribution with mean $2$. But why doesn't the first method work?

Please help! I'd really appreciate it.

  • 2
    $\begingroup$ In general, E[X^2] != E[X]^2 $\endgroup$
    – Steve S
    Oct 26, 2014 at 18:50
  • $\begingroup$ But they are equal if X is independent, correct? $\endgroup$
    – Lindsey
    Oct 26, 2014 at 19:21
  • $\begingroup$ X can't just be 'independent'. Independence is a property of a set of random variables. You can use independence of X and Y (notice that that is independence with respect to both the variables) to say that E[XY] = E[X]E[Y]. But it doesn't make sense to say that X is "independent" because the natural next question would be "independent of what?". It can't be independent of itself which is what you would need to say that E[X^2] = E[X*X] = E[X]E[X]. $\endgroup$
    – Dason
    Oct 26, 2014 at 19:30
  • 4
    $\begingroup$ Recall the formula for the variance of X: Var(X) = E[X^2] - E[X]^2. Now, the only way that E[X^2] would be equal to E[X]^2 is if X does not vary at all (i.e. it's a constant). $\endgroup$
    – Steve S
    Oct 26, 2014 at 19:33
  • 1
    $\begingroup$ Okay, I see now that was a pretty silly question. Sorry! But thank you all so much (especially @Dason for completely addressing my issue). $\endgroup$
    – Lindsey
    Oct 27, 2014 at 1:09

2 Answers 2


why can't I expand the $(X + Y)^2$ to get $E(X^2 + 2XY + Y^2)$

You can.

and then use linearity to get $E(X^2) + 2E(X)E(Y) + E(Y^2)$

You can

Then since $X$ and $Y$ are independent,

right ... but you already used independence in the previous step when you wrote $E(XY)$ as $E(X)\,E(Y)$.

this would give $E(X)^2 + 2(1)(1) + E(Y)^2 = 1 + 2 + 1 = 4$.

Nope. You just went $E(X^2) = E(X)^2$; that's not true. You got this manipulation right when you did it the other way. Note that:

$E(X^2) = \text{Var}(X) + E(X)^2 = 1 + 1 = 2$

and then the result follows.

(It would be easiest to add the random variables first and then find the expectation of the square, but it's quite doable by expanding the square)


You cannot say $E(X^2)=E(X)^2$ since $X$ is not independent of $X$. For this problem you can set $Z=X_1+X_2 \sim \text{Poisson}$ with mean $2$. Than you can find $E(Z^2)$ easily.


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