# Variance of integral of Poisson variables

I have a stochastic quantity (not sure if it is a proper stochastic process), defined as follows:

$$I = \int d x f(x) X(x)$$ $$f(x)$$ is a positive function of real variable, defined over the integral domain. $$X(x)$$ is a sequence of random variables, defined over $$x$$ as their parameter space.

In particular, their are uncorrelated, Poisson distributed, with means and variances equal to $$\mu_{X}(x)$$. I'm interested in the mean and variance of $$I$$.

Using the first and second order moments I get

$$\text{E}[I] = \text{E}\left[\int d x f(x) X(x)\right]=\int d x f(x)\mu_{X}(x)$$ and \begin{align} \text{E}[I^2] =& \text{E}\left[\int d x f(x) X(x)\int d y f(y) X(y)\right] \\ =& \int d x d y f(x) f(y) \text{E}\left[X(x)X(y)\right] \\ =& \int d x d y f(x) f(y) \delta(x-y)\text{E}\left[X(x)^2\right]\\ =& \int d x f(x)^2 \text{E}\left[X(x)^2\right]\\ =& \int d x f(x)^2 \text{Var}\left[X(x)\right] + \text{E}\left[X(x)\right]^2 \\ =& \int d x f(x)^2 (\mu_{X}(x) +\mu_{X}(x)^2) \end{align} Although the variance of $$I$$

\begin{align} \text{Var}[I] =& \text{E}[I^2]-\text{E}[I]^2 \\ =& \int d x f(x)^2 (\mu_{X}(x) +\mu_{X}(x)^2)- \left(\int d x f(x)\mu_{X}(x)\right)^2 \end{align} is not necessarily positive.

Consider, e.g. $$f(x) = 1$$, $$\mu_{X}(x)=x$$, and $$I$$ to be defined over $$[1,3]$$

\begin{align} \text{Var}[I] =& \int_1^3 d x (x+x^2)- \left(\int_1^3 d x x\right)^2\\ =& - \frac{10}{3} \end{align} Any help figuring out the issue would be appreciated. A somewhat related answer is here although it actually goes in the same direction as my conclusion.

• I'm not sure exactly what's meant by $\delta$ here. I agree with everything before that line and disagree with the following line, so I think that's where the mistake is. May 8, 2023 at 8:45
• @JoeMansley, thanks for your request of clarification. The random variables defined by X(x) are uncorrelated, so the joint expectectation value in the last line you agree with is non-zero only when the two integration variables take the same value (i.e. x=y). That's why I am using in the first line you disagree with a Dirac delta for x-y and replacing the joint expectation value with the second order moment of X(x). Do you still disagree with my derivation? Jul 30, 2023 at 22:13
• I don't agree that the joint expectation is necessarily $0$ when $x\neq y$ Aug 1, 2023 at 9:09
• I may have been unclearin my question: I assume that the sequence variables X(x), are uncorrelated, which if I am not mistaken means by definition: $$x \neq y \Rightarrow E[X(x) X(y)] = 0$$ Aug 1, 2023 at 14:09
• I believe you are mistaken. That they are uncorrelated means: $x \neq y \Rightarrow E[X(x)X(y)]=E[X(x)]E[X(y)]$ Aug 1, 2023 at 18:45

Thanks to @JoeMansley, the derivation was flawed. Here follows what I believe is the correct one, and doesn't lead to incorrect results (i.e. a negative variance).

\begin{align} \text{E}[I^2] =& \text{E}\left[\int d x f(x) X(x)\int d y f(y) X(y)\right] \\ =& \int d x d y f(x) f(y) \text{E}\left[X(x)X(y)\right] \\ =& \left(\int d x f(x) \mu_{X}(x)\right)^2 + \int d x f(x)^2 \text{Var} \left[X(x)^2\right]\\ =& E\left[I\right]^2 + \int d x f(x)^2 \text{Var}\left[X(x)\right] \end{align}

Therefore the variance of $$I$$ reads:

$$\sigma^2_I = \int d x f(x)^2 \sigma_X^2(x)$$

and for the special case at hand of Poisson, uncorrelated variables, I have:

$$\sigma^2_I = \int d x f(x)^2 \mu_X(x)^2$$

which is a generalized result to an infinite sum of uncorrelated variables of

$$\sigma^2_{aX+bY} = a^2\sigma^2_X + b^2 \sigma_Y^2$$