I stumbled upon problem 254 from the SOA Exam P list in


for which I am puzzled by the solution described in


Basically, the way they solved it is by stating that the second moment of the weighted average of Poisson variables, is the weighted average of the second moments. For the first moment (expected value) this is obviously true in general, but for the second moment one expects

$E[(aX+bY)^2] = a^2 E[X^2] + 2abE[XY] + b^2 E[Y^2]$.

There is no mention of indepence or correlation at all, only that these are Poisson variables and their means are positive. Would that be suficient to state that

$E[(aX+bY)^2] = a E[X^2] + bE[Y^2]$

as the solution claims?

New contributor
XaviZ is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
  • 1
    $\begingroup$ (1) Your last formula is not what the solution claims.--and it's false. Consider, for instance, the situation where $a$ and $b$ are negative. No matter what $X$ and $Y$ might be, the left hand side is nonnegative but the right hand side is non-positive. (2) The statement in question is formulated and proven at stats.stackexchange.com/a/16609/919. (3) Expectation is linear, regardless of any correlation among the variables. $\endgroup$
    – whuber
    Commented Jul 10 at 20:54


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.