Conditional expectation of the sum of two random variables

Question

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

Answer 1

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 $k\ge 0:$ $$E[N_1(t)+N_2(t) \mid N_2(t)=k] = \lambda_1t +k $$

Answer 2

For every independent integrable random variables $X$ and $Y$, $$E(X+Y\mid Y)=E(X)+Y,$$ by linearity of conditional expectation, hence $$E(X+Y\mid Y=y)=E(X)+y.$$

Answer 3

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