# Variance of discrete probability distribution

I was wondering how I should calculate the variance of the following discrete probability distribution:

$$P(y = 0|X) = w + (1-w)e^{-\mu}$$ $$P(y = j|X) = (1-w)e^{-\mu}\mu^{y}/y! \qquad j=1,2...$$

The variance is supposed to be:

$$Var(y|w, \mu) = (1-w)[\mu + w\mu^{2}]$$

I suppose the total variance is the sum of the variance associated to the case $y=0$ and $y=j$ for $j>0$. Correct me if I'm wrong, but I don't think something like $Var(X + Y) = Var(X)+Var(Y)+Cov(X,Y)$ is appropriate because $y$ is a single random variable which takes several values with distinct probability distributions. However, the $y=0$ case doesn't depend on $y$, so $Var(y=0|w,\mu)$ should be zero. The case $y=j$ could be something like:

$$\sum_{1}^{\infty} (y-(\mu-w \mu))^{2}(1-w)e^{-\mu}\mu^{y}/y!$$

The part $\mu - w\mu$ is due to the expected value of the probability distribution $P(y=j|X)$:

$$\sum_{1}^{\infty} y(1-w)e^{-\mu}\mu^{y}/y!=\mu - w\mu$$

This expected value is correct. However, the value of the variance calculated as I described above, is not correct. It gives a much more complicated expression. Interestingly, summing from $0$ to $\infty$, produces a very similar result to the correct answer:

$$\sum_{0}^{\infty} (y-(\mu-w \mu))^{2}(1-w)e^{-\mu}\mu^{y}/y!=(1-w)\mu(1+w^{2}\mu)$$

By the way, I'm using Mathematica to obtain these results. I'm only interested in the correct approach to obtain this kind of variance, not in the details of the calculation.

• as an aside: What is the relevance of the conditional $X$ in your opening equations. There is no $X$ on the RHS? – wolfies Nov 8 '14 at 6:40
• The reason we have $X$ is because $w$ depends on $X$ through a logit regression. That's not relevant in this particular question, though. – Robert Smith Nov 8 '14 at 7:48

This is known as a Zero-inflated Poisson model (or ZIP). Here, $$Y$$ has pmf $$f(y)$$:

Then, $$Var(Y)$$ is:

where I am using the Var function from the mathStatica package for Mathematica.

As to how to calculate the variance:

There are two standard approaches to calculating variance:

1. $$Var(Y) = E[Y^2] - (E[Y])^2$$
2. $$Var(Y) = E\big[(Y- E[Y])^2\big]$$

If you use the FIRST approach, $$Var(Y) = E[Y^2] - E[Y]^2$$ ... note that the first and second moments are calculated as of form:

$$\sum _{y=0}^{\infty } (y f) \quad \text{and} \quad \sum _{y=0}^{\infty } (y^2 f)$$

Note that when $$Y=0$$, the expression $$y^i f$$ = 0, so the $$Y = 0$$ line of your piecewise pmf contributes 0 to both the first and second moment, and accordingly you can calculate both excluding the $$Y= 0$$ line, and simply summing from $$y = 1$$ to $$\infty$$.

In Mathematica (which you are using), you would enter:

Sum[y^2 f, {y, 0, Infinity}] - Sum[y f, {y, 0, Infinity}]^2   // Simplify


Summing from 1 yields the same outcome:

Sum[y^2 f, {y, 1, Infinity}] - Sum[y f, {y, 1, Infinity}]^2   // Simplify


However, if you use the SECOND APPROACH $$Var(Y) = E\big[(Y- E[Y])^2\big]$$ to calculating variance, then you CANNOT exclude the Y = 0 case in the summation because $$(Y- E[Y])f$$ is NOT equal to 0 when $$Y = 0$$.

• Thank you. You're correct. I think a simpler way to look at it is that $E[(Y-E[Y])^{2}]$ should be $\sum_{0}^{\infty}(y-(\mu-w\mu))^{2}p(y)$. Therefore, the first term for $y=0$ is accompanied with $p(y=0)$ as in $(\mu-w\mu)p(y=0) = (\mu-w\mu)^{2}(w+(1-w)e^{-\mu})$ and the following terms are what we already had: $\sum_{y=1}^{\infty}(y-(\mu -w\mu))^{2}(1-w)e^{-\mu}\mu^{y}/y!$, which gives the correct result. – Robert Smith Nov 8 '14 at 7:45