Conditional expectation of the sum of two random variables I've got some difficulties in calculating the conditional expectation of the sum of two RV. 

I am not sure if I correctly formalized the scenario I am looking at. So I am trying to describe it first: I have two independent Poisson processes $P_1$ and $P_2$ with specific means $\lambda_x$ and $\lambda_y$. They are always executed one after the other. So $n$ runs of an experiment involve $n$ runs of both $P_1$ and $P_2$. I know that $E[N|T=t] = \lambda\,t$. What I'd like to calculate is the expectation that they sum of $n$ consecutive runs of the experiment given that the sum of one of the processes is $k$, i.e. $P_2=k$.
I've tried to formalize this into the following:

Let $X$ and $Y$ be two independent, exponentially distributed random variables with means $\lambda_x$ and $\lambda_y$. Let $Z$ be the sum of the two variables, i.e. $Z=X+Y$. So $Z$ should also be exponentially distributed, hence $Z\sim P(\lambda_x+\lambda_y)$.

I'd like to calculate the conditional expectation of $Z$, given that $Y=k$.

I have $P(Z=n| Y=k) = \frac{P(X+Y=n, Y=k)}{P(Y=k)} = \frac{P(X=n-k)P(Y=k)}{P(Y=k)}$. But this would cancel out the last term which doesn't seem to be right.

Any help is appreciated.

Thanks
The notation (or lack thereof) is confusing and what (I think) you are asking for makes no sense but... Let $\{N_1(t):t\ge 0\}, \{N_2(t);t \ge 0\}  $ be independent Poisson processes with rates $\lambda_1,\lambda_2.$ For a given time $t_0$ we observe values of $N_1(t_0), N_2(t_0)$ and then repeat $n$ times and sum the $n $ observed values for each process. That is the same as observing the values $N_1(nt_0), N_2(nt_0).$ The comments about one after another and consecutive runs do not matter.

Now let $t$ be any fixed time, possibly $t=nt_0,$ and for any $k\ge 0:$ 
$$E[N_1(t)+N_2(t) \mid N_2(t)=k] = \lambda_1t +k $$
The answer of Mr.Spot seems to bring me closer to the right answer, I believe. I again assume that $\lambda_1$ and $\lambda_2$ are the means and $n$ the number of repetitions. So the sum of the $n$ observed values would be $\lambda_1n$ and $\lambda_2n$.
Starting with what we had at the end:

$E[N_1(t)+N_2(t) \mid N_2(t)=k] = \lambda_1n + \lambda_2n $ and $k=\lambda_2n$.

We see that $n=\frac{k}{\lambda_2}$. Hence, we can write:

$E[N_1(t)+N_2(t) \mid N_2(t)=k] = \lambda_1 \frac{k}{\lambda_2} + k = k(\frac{\lambda_1}{\lambda_2}+1)$

However, filling in some values for $k, \lambda_1$ and $\lambda_2$ doesn't yield the expected results that I got from a simulation of the experiment. However, 
$E[N_1(t)+N_2(t) \mid N_2(t)=k] = k(\frac{\lambda_2}{\lambda_1}+1)$ does work. Note its $\lambda_2$ over $\lambda_1$. But I don't know how to obtain it.

Thanks