Mean, Variance, and Conditional PMF from Poisson Process

Question

In good years, quarrels between Tolstoy and his wife occurred according to a Poisson process with rate λ = 5 per month. In bad years, it was a Poisson process with rate μ = 10 per month. Suppose each year was equally likely to be a good or a bad year independently of what happened in previous years.

(a) Compute the mean and variance of the total number of quarrels in a randomly selected year.

(b) Suppose there were a total of 12 quarrels in January 1891. Compute the conditional probability mass function of the total number of quarrels in the year 1891 given this information. Hint: was this a good year or a bad year?

Here's my approach so far:

G = next year is a good year B = next year is a bad year N(t) = number of quarrels before time (month) t

$P(G) = P(B) = 0.5$

$N(t)|G \sim Pois(5t)$

$N(t)|B \sim Pois(10t)$

Is it correct to say:

$$P(N(t) = n) = P(N(t) = n | G)P(G) + P(N(t) = n | B)P(B)$$

$$P(N(t) = n) = \dfrac{(5t)^ne^{-5t}}{n!}0.5 + \dfrac{(10t)^ne^{-10t}}{n!}0.5$$

$$E[N(t=12)] = \sum_{i=0}^{\infty}n[\dfrac{(5t)^ne^{-5t}}{n!}0.5 + \dfrac{(10t)^ne^{-10t}}{n!}0.5]$$

and obtaining the variance would be a similar process?

Answer 1

Sure you can do that.

You are using the Laws of Total Expectation and Variance. Also remember that you (should) know the mean and variance of a Poisson distribution each equal the rate parameter.

$$\begin{align}\mathsf E(N(t)\mid G)&=\sum_{n=0}^\infty \bigl(n~(5t)^n\mathrm e^{−5t}/n!\bigr) \\ &= 5t\\\mathsf{Var}(N(t)\mid B) &=\sum_{n=0}^\infty \bigl((n-10t)^2~(10t)^2\mathrm e^{-10t}/n!\bigl)\\&= 10t\\\ddots&\quad\text{et cetera}\end{align}$$

Let $Y$ be the indicator random variable that the year is good. $G:=\{Y=1\}, B:=\{Y=0\}$

$$\begin{align} Y&\sim\mathcal{Bern}(0.5)\\ N(t)\mid Y=1 &\sim\mathcal{Pois}(5t)\\ N(t)\mid Y=0&\sim\mathcal{Pois}(10t)\\[3ex]\mathsf E(N(t)\mid Y) &= 5t\mathbf 1_{Y=1}+10t\mathbf 1_{Y=0}\\\mathsf{Var}(N(t)\mid Y) &= 5t\mathbf 1_{Y=1}+10t\mathbf 1_{Y=0}\\[3ex]\mathsf E(5t\mathbf 1_{Y=1}+10t\mathbf 1_{Y=0}) &= 5t\mathsf P(Y\,{=}\,1)+10t\mathsf P(Y\,{=}\,0) \\[1ex] \mathsf{Var}(5t\mathbf 1_{Y=1}+10t\mathbf 1_{Y=0}) &=(25t^2\mathsf P(Y\,{=}\,1)+100t^2\mathsf P(Y\,{=}\,0))-(5t\mathsf P(Y\,{=}\,1)+10t\mathsf P(Y\,{=}\,0))^2\\[3ex]\mathsf E(N(t)) &= \mathsf E(\mathsf E(N(t)\mid Y))\\\mathsf {Var}(N(t))&= \mathsf E(\mathsf{Var}(N(t)\mid Y))+\mathsf{Var}(\mathsf E(N(t)\mid Y))\end{align}$$

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For (b) Let $N_J$ be count for the month of January, and $N_R$ be the count for the remaining 11 months of that year.   You therefore seek $\mathsf P(N_J\,{+}\,N_R\,{=}\,n\mid N_J\,{=}\,12)$.

$$\begin{align} \mathsf P(N_J\,{+}\,N_R\,{=}\,n\mid N_J\,{=}\,12)&~=~\dfrac{\mathsf P(N_J\,{=}\,12, N_R\,{=}\,n\,{-}\,12)}{\mathsf P(N_J\,{=}\,12)}\end{align}$$

For a given type of year, the counts $N_J, N_R$ are conditionally independent since they occur over disjoint intervals. Apply the Law of Total Probability to the denominator and numerator.

$${N_J\mid G\sim\mathcal{Pois}(5)\\ N_J\mid B\sim\mathcal{Pois}(10)\\N_R\mid G\sim\mathcal{Pois}(55)\\ N_R\mid B\sim\mathcal{Pois}(110)\\ N_J\perp N_R\mid G\\N_J\perp N_R\mid B}$$

Answer 2

My approach in $(b)$ is slightly different from Graham Kemp's; kindly indulge in my alternate approach.

Let $D$ be the event that there are exactly $12$ quarrels observed in $1/1891$, and take $X$ at the total numbers of quarrels in the year $1891$.

The conditional pmf of $X$ given $D$, namely $p_{X|D}$, is supported on $\{12,13,14,...\}$ and can be derived by the total law of conditional probability.

For $x\in \{12,13,14,...\}$ we have $$\begin{eqnarray*}p_{X|D}(x)&=&\mathbb{P}(X=x|D) \\ &=& \mathbb{P}(X=x|D,G)\mathbb{P}(G|D)+\mathbb{P}(X=x|D,B)\mathbb{P}(B|D) \\ &=& \mathbb{P}(X=x|D,G)\times \frac{\mathbb{P}(D|G)\mathbb{P}(G)}{\mathbb{P}(D|G)\mathbb{P}(G)+\mathbb{P}(D|B)\mathbb{P}(B)}+\mathbb{P}(X=x|D,B)\times \frac{\mathbb{P}(D|B)\mathbb{P}(B)}{\mathbb{P}(D|G)\mathbb{P}(G)+\mathbb{P}(D|B)\mathbb{P}(B)}\end{eqnarray*}$$ By using the facts that $$\mathbb{P}(X=x|D,G)=e^{-55}\times \frac{55^{x-12}}{(x-12)!}$$ $$\mathbb{P}(X=x|D,B)=e^{-110}\times \frac{110^{x-12}}{(x-12)!}$$ $$\mathbb{P}(D|G)=e^{-5}\times \frac{5^{12}}{12!}$$ $$\mathbb{P}(D|B)=e^{-10}\times \frac{10^{12}}{12!}$$ $$\mathbb{P}(G)=\mathbb{P}(B)=\frac{1}{2}$$ you can obtain your formula for $p_{X|D}(x)$.