Solved – Variance of the sum of a Poisson-distributed random number of (normally distributed) random variables

mathematical-statisticsmeannormal distributionpoisson distributionvariance

Suppose I have a fruit stand where I sell oranges. Every day I have an average of $\lambda$ people (Poisson distributed) each buying a normally distributed weight quantity of oranges with mean $\mu$ and standard deviation $\sigma$.

What's the mean and variance of the number of pounds of oranges I sell per day?

OK, the mean is trivial: $\lambda \cdot \mu$. But how do I calculate the variance?

Strangely enough I've come up with an answer through trial and error that seems to fit really well with my Monte Carlo simulations, but I can't convince myself that it's actually the correct answer in theory. If someone can point me to the "right" answer, I'll post the one I got empirically for comparison.

Best Answer

This is a compound Poisson distribution

The law of total variance gets you the answer.

Note that this answer doesn't rely on the normality of the $X$'s at all; it applies to any distribution with mean $\mu$ and variance $\sigma^2$.

In the following, $N$ is the Poisson r.v., $X$'s are individual components, $Y$ is the sum of components, $Y|N$ is the sum of components given a specific count of terms, $N$:

$$\text {Var}_{Y}(Y)=E_{N}\left[\text {Var}_{{Y|N}}(Y)\right]+\text {Var}_{N}\left[E_{{Y|N}}(Y)\right]$$

Substituting in:

$$=\text {E}_{N}\left[N\text {Var}_{X}(X)\right]+\text {Var}_{N}\left[N\text {E}_{X}(X)\right]\,$$

$$=\text {E}_{N}\left[N\sigma^2\right]+\text {Var}_{N}\left[N \mu\right]\,$$

$$=\sigma^2\text {E}_{N}\left[N\right]+\mu^2\text {Var}_{N}\left[N \right]\,$$

$$=\sigma^2\lambda+\mu^2\lambda$$

$$=(\sigma^2+\mu^2)\lambda$$

(... which is also $\lambda E(X^2)$)

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Solved – Standard deviation of the sum of two normally distributed random variables

It's interesting that you are trying to use the matrix form when this is more of a elementary problem.

Remember $$ \mbox{Var}(X + Y) = \mbox{Var}(X) + \mbox{Var}(Y) + 2 \mbox{Cov}(X,Y). $$

You don't say anything about the relation between $X$ and $Y$, so either you assume a covariance of zero or the problem is not solvable due to missing informations.

The matrix form of this is $$ \mbox{Cov}(AX) = A \cdot \mbox{Cov}(X)\cdot A'. $$

This covariance matrix seems to be your $\Sigma$. I don't know why you inverted it in your formula. Perhaps I misunderstand your equation? You know that $\Sigma_{1,1} = 6$ and $\Sigma_{2,2} = 8$. These are just the variances of $X$ and $Y$ respectively. Again, since you do not have a covariance and since you cannot calculate it from the information given, the question is either unsolvable or you have have to assume independence, which implies $\Sigma_{1,2}=\Sigma_{2,1}=0$. This will then give the same result as the equation for $\mbox{Var}(X+Y)$. Remember to take the root to get the standard deviations.

Solved – Calculating the probability of a sum of Poisson-distributed random variables

If you assume that customers demands are independant and Poisson, the total demand will be a Poisson distribution as well with parameter equal to the sum of the individual Poisson parameters (see wikipedia for example).

A C++ code example (I am not so familiar with Java):

double cumulative(double x, User * users, int nusers) 
{
    double lambda = 0.;
    for (unsigned i = 0; i < nusers; ++i) lambda += users[i].mean();
    return PoissonDistribution(lambda).cumulativeProbability(x);
}

Best Answer

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Solved – Calculating the probability of a sum of Poisson-distributed random variables

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