Problem: Consider a Poisson process with parameter $\lambda$ . Let $T$ be a random variable representing the time required to observe the first event, and $X_\frac{T}{K}:=N\left(\frac{T}{K}\right)$ be the random variable representing the number of events in the next $\frac{T}{K}$ units of time. Show that $E\left\{N\left(\frac{T}{K}\right) T \right\}=\frac{2}{\lambda K}$ and $E\left\{\left[N\left(\frac{T}{K}\right) T \right]^2\right\}=\frac{6}{\lambda^2 K} + \frac{24}{\lambda^2 K^2}$
Attempt at the Solution: I thought of solving this problem using following fact from probability theory: $E(XY)= Cov(X,Y) + E(X)E(Y)$ where I take $X$ to be $N\left(\frac{T}{K}\right)$ and $Y$ to be $T$. The tricky part is finding $Cov(X,Y)$ and that's where I'm stuck:
\begin{align*} Cov(X,T)=E[(X-\mu_X)(T-\mu_T)] &= \int^\infty_0 \sum^\infty_{x=0} (x-\mu_X)(t-\mu_T)f_X f_T \,dT \\ &= \int^\infty_0 \left\{ \sum^\infty_{x=0} (x-\mu_X)f_X \right\}(t-\mu_T) f_T \,dT \\ \end{align*} where $f_{X_\frac{T}{K}}(x)= \frac{e^{-\frac{\lambda T}{K}\left(\frac{\lambda T}{K}\right)^x}}{x!}$ and $f_T(t)= \lambda e^{-\lambda t}$.
Since \begin{align*} \sum^\infty_{x=0} (x-\mu_X)f_X &= \sum^\infty_{x=0} \left(x-\frac{\lambda T}{K}\right)\frac{e^{-\frac{\lambda T}{K}\left(\frac{\lambda T}{K}\right)^x}}{x!} \\ &= \sum^\infty_{x=0} x \frac{e^{-\frac{\lambda T}{K}\left(\frac{\lambda T}{K}\right)^x}}{x!} - \sum^\infty_{x=0} \frac{\lambda T}{K} \frac{e^{-\frac{\lambda T}{K}\left(\frac{\lambda T}{K}\right)^x}}{x!} \\ &= \mu_X - \frac{\lambda T}{K} \\ &= \mu_X - \mu_X \\ &= 0 \end{align*} $Cov(X,T)=0$ and hence, $E(XT)=E(X)E(T)=\left(\frac{\lambda T}{K}\right)\left(\frac{1}{\lambda}\right)=\frac{T}{K}$
What am I missing here? Is there an alternative method to solve this?
Use a conditional expectation argument here. Recall that $E[T^2] = \frac{2}{\lambda^2}$ and write $$ E\left\{ N\left(\frac{T}{K} \right) T\right\} = E\left\{ E \left[ N\left(\frac{T}{K} \right) \mid T \right] T\right\} = E\left\{ \left(\frac{\lambda T}{K} \right) T \right\} = \frac{\lambda}{K} E[ T^2 ] = \frac{2}{\lambda K}. $$