Customers arrive at the shopping mall according to a Poisson process with $\lambda = 3$ and the process begins at $t = 0$. Let $X$ be the number of arrivals in the interval $[2, 4]$. (i) Find $P(X \leq 4|X \geq 2)$. (ii) Let $Y$ be the time of the $10$th arrival. Find the values of $m$ and $n$ if $P(Y > 8) = \sum_{j = 0}^{m} \frac{n^j \cdot e^{-n}}{j!}$ . (iii) Let $W$ be the time of the third arrival. Find $E[\frac{1}{W^2}] $.
My attempt:
(i) $X$ follows a Poisson distribution with $\lambda = 3 *(4 - 2) = 6$. Hence
$P(X \leq 4|X \geq 2) = \frac{P(2 \leq X \leq 4)}{1 -P(X < 2)} = \frac{P(X = 2) + P(X = 3) + P(X = 4)}{1 - P(X = 0) - P(X = 1)} \approx 0.27243228$
(ii) Consider the following argument:
Let $T$ be the time of the $k$th arrival where $k \in \mathbb{Z}^+$. If $S$ is the number of arrivals in $[0, s]$ then $S$ is Poisson distributed with $s\lambda$. So
$P(T \leq t) = P(S \geq k) = 1 - P(S < k) = 1 - \sum_{j = 0}^{k - 1} \frac{e^{-t\lambda}(t\lambda)^j}{j!}$
By a similar argument, if $X$ is the number of arrivals in $[2, 4]$ then this is equivalent to the number of arrivals in $[0, 2]$.
$P(Y > 8) = 1 - P(Y \leq 8) = P(X < 10) = \sum_{j = 0}^{9} \frac{e^{-8 \cdot 6}(8 \cdot 6)^j}{j!}$
So $m = 9, n = 48$.
(iii) No idea how to do this one.
For the ones I attempted, are they correct? The second part seems incorrect and I have no clue about the third part. Any assistance much appreciated.
Solutions to the first two parts look correct to me.
For the third part, if $W_n$ denotes the time of the $n^\mathrm{th}$ arrival, then $W_n$ is distributed according to $\mathrm{Gamma}(n,\lambda)$. Hence the pdf of $W_n$ is given by \begin{align*} f_{W_n}(t)=\frac{\lambda\exp(-\lambda t)(\lambda t)^{n-1}}{(n-1)!} \end{align*} Since we want to know about $W_3$, its pdf is given by \begin{align*} f(t)=\frac{\lambda^3\exp(-\lambda t)t^{2}}{2} \end{align*} Then, \begin{align*} \mathbb{E}\left(\frac{1}{W_3^2}\right)&=\int_{0}^\infty dt\frac{1}{t^2}f(t)\\ &=\frac{\lambda^3}{2}\int_{0}^\infty dte^{-\lambda t}\\ &=\frac{\lambda^2}{2} \end{align*}