Find the density function, pdf, of $Y$, the arrival time of the $k$th event of a poisson process. I am having difficulty in approaching the following problem:
Suppose $X \sim \text{Poisson}(\lambda)$ and let $k$ exist $N$ be fixed. Also, let $Y$ be the (random) arrival time of the $k$th Poisson event (from the distribution of $X$). Find the density function of $Y$.
Is it asking me to incorporate the Poisson PDF and show the derivation for some random arrival time of $Y$ in a $k$th poisson event?
I know what the Poisson PDF is, but not sure how I would apply this.
 A: As stated in the problem we let $Y_k$ be the time of the $k$-th event of the poisson process $X(t)$ with rate $\lambda$.
$Y_k = Z_1 + Z_2 + \cdots + Z_k$ 
where $Z_k,\: k=1,2...$ are the interarrival times.
It is a straight forward exercise to show that $Z_k$ are iid exponential random variables with parameter $\lambda$. Note that an exponential random variable with parameter $\lambda$ is a gamma random variable with parameters $(1,\lambda)$. 
It is another straight forward excercise to show, by induction, that the sum $Y_k$ is a gamma random variable with parameter $(k, \lambda)$.
Hence $Y_k$ has a pdf given by 
$$ f_{Y_k}(t) =
\begin{cases}
\lambda e^{-\lambda t} \frac{(\lambda t)^{k-1}}{(k-1)!},  & t > 0 \\
0, & t < 0
\end{cases}
$$

You can show that the $Z_k's$ are iid exponentials like this:
$$P(Z_1 > t) = P\{X(t) = 0\} = e^{-\lambda t}$$
$$F_{Z_1}(t) = P(Z_1 \leq t) = 1-e^{-\lambda t}$$
Hence, $Z_1$ is an exponential r.v.
Let $f_1(t)$ be the pdf of $Z_1$.
Then we have that
$$
\begin{align}
P(Z_2 > t) & = \int P(Z_2 > t | Z_1 = \tau)f_1(\tau) d \tau \\
 &   \\ 
 & = \int P[X(t + \tau) - X(\tau)=0]f_1(\tau) d \tau \\
 &   \\ 
 & = e^{-\lambda t}\int f_1(\tau) d \tau \\
 &   \\
 & =  e^{-\lambda t}
\end{align}
$$
which indicates that $Z_2$ is also an exponential r.v. with parameter $\lambda$ and is independent of $Z_1$. Repeating the same argument, you can conclude that $Z_1, Z_2, ....$ are iid exponential r.v.'s with parameter $\lambda$.
