$N$ is a Poisson distribution with mean $4$. Find $\operatorname{Var}(N\mid N \geq 4)$

Question

You are given that $N$ has Poisson distribution with mean $4$. Find $\operatorname{Var}(N\mid N \geq 4)$

I tried to use the definition of variance, where $\operatorname{Var}(X) = E(X^2) - E(X)^2$

Then $ P(N\mid N \geq4) = 1-P(N <4) = 1 - \dfrac{71}{3}e^{-4}$

I'm stuck now about how to use the formula above to calculation the conditional expected value.

Any help would be appreciated.

Answer 1

$\newcommand{\var}{\operatorname{var}}\newcommand{\E}{\operatorname{E}}$Let $Y=0\text{ or }1$ according as $N<4$ or $N\ge 4$. For the Poisson distribution the expectation is the same as the variance, so we have $$ 4 = \var(N) = \var(\E(N\mid Y)) + \E(\var(N\mid Y)). $$ Next $$ 4 = \E(N) = \E(N\mid N < 4)\Pr(N < 4) + \E(N\mid N \ge 4)\Pr(N\ge 4). \tag 1 $$ It is not hard to find $\Pr(N<4)$, and in order to find $\E(N\mid N<4)$ you need four conditional probabilities: \begin{align} \Pr(N=0\mid N<4) & = \frac{e^{-4}}{e^{-4} + 4e^{-4} + \frac{16e^{-4}} 2 + \frac{64 e^{-4}} 6} = \frac 1 {1+4+\frac{16} 2 + \frac{64} 6} = \frac 3{71} \\[8pt] \Pr(N=1 \mid N<4) & = \frac{12}{71} \\[8pt] \Pr(N=2 \mid N<4) & = \frac{24}{71} \\[8pt] \Pr(N=3 \mid N<4) & = \frac{32}{71} \end{align} So $\E(N\mid N<3) = 156/71 \approx 2.197\ldots$

$\Pr(N<4)$ is readily found and $\Pr(N\ge 4)$ is $1$ minus that.

This makes it possible to find $\E(N\mid N\ge 4)$.

Then we have the random variable $\E(N\mid Y)$: $$ \E(N\mid Y) = \begin{cases} \E(N\mid N<4) & \text{with the probability $p$ found above}, \\ \E(N\mid N\ge 4) & \text{with probablity }1-p. \end{cases} $$ Hence $$ \var(E(N\mid Y)) = \Big( \E(N\mid N<4) - \E(N\mid N\ge 4) \Big)^2 p(1-p). $$

Proceed similarly to find the other term in $(1)$.

Answer 2

Let $$M = N \mid N \ge 4.$$ That is to say, $$\Pr[M = k] = \frac{\Pr[N = k]}{\Pr[N \ge 4]}, \quad k = 4, 5, \ldots.$$ Then $$\operatorname{Var}[M] = \operatorname{E}[M^2] - \operatorname{E}[M]^2 = \sum_{k=4}^\infty k^2 \frac{\Pr[N = k]}{\Pr[N \ge 4]} - \left(\sum_{k=4}^\infty k \frac{\Pr[N = k]}{\Pr[N \ge 4]} \right)^2.$$ This is just one way to do the calculation; it may not necessarily be the most elegant.