I have to calculate the following conditional expectations for a homework. Please help me !
Let be $\{X_{t}: t \geq 0 \}$ a Poisson Process with the parameter $\lambda=2$ . Let be $W_{n}$ the instant at which the nth event occurs. Furthermore, you know that $W_{n}=T_{1}+ \ldots+T_{n}$, where $T_{n}\sim Exp(\lambda), \forall n \in \mathcal{N} $ represents the times when the events occurred.
So, $W_{n} \sim Gamma(n,\lambda)$. But I don't know how to calculate this:
- $E(W_{7}|X_{2}=3)=?$
- $E(X_{7}|W_{5}=4)=?$
I only know that $E(W_{7})= \frac{7}{2}$ and that the exponential distribution has the property of memory loss.
There may be an easier route to the solution, but this is my approach.
Let $f_{W_3W_4W_4}(x,y,z)$ denote the joint density of $(W_3,W_4,W_7)$. Note how $$\begin{eqnarray*}f_{W_3W_4W_7}(x,y,z)&=&f_{W_7|W_3W_4}(z|x,y)f_{W_4|W_3}(y|x)f_{W_3}(x)\\ &=& f_{W_7|W_4}(z|y)f_{W_4|W_3}(y|x)f_{W_3}(x)\end{eqnarray*}$$ The conditional distribution of $W_7$ given that the fourth arrival occurs at time $y$ is sum of $y$ with an $\text{Erlang}(k=3,\lambda=2)$ random variable. Symbolically, $$W_7|\{W_4=y\}= A+y \text{ where } A\sim\text{Erlang}(k=3,\lambda=2)$$ Similarly, the conditional distribution of $W_4$ given that the third arrival occurs at time $x$ is the sum of $x$ with an $\text{Exponential}(\lambda =2)$ random variable. Symbolically, $$W_4|\{W_3=x\}= B+x \text{ where } B\sim \text{Exponential}(\lambda =2)$$ We also know $W_3 \sim \text{Erlang}(k=3,\lambda =2)$ so for any $(x,y,z)$ such that $0\leq x\leq y \leq z $ we have $$\begin{eqnarray*}f_{W_3W_4W_7}(x,y,z)&=&\Big[4(z-y)^2e^{-2(z-y)}\Big]\cdot \Big[2e^{-2(y-x)}\Big]\cdot \Big[4x^2e^{-2x}\Big] \\ &=& 32x^2(z-y)^2e^{-2z}\end{eqnarray*}$$ Since $\{X_2=3\}$ is the same as $\{W_3 \leq 2,W_4>2\}$ we get $$\begin{eqnarray*}\mathbb{E}(W_7|X_2=3) &=& \mathbb{E}(W_7|W_3 \leq 2,W_4>2) \\ &=& \frac{\int_0^2 \int_{2}^{\infty} \int_y^{\infty}zf_{W_3W_4W_7}(x,y,z)\mathrm{d}z\mathrm{d}y\mathrm{d}x}{\int_0^2 \int_{2}^{\infty} \int_y^{\infty}f_{W_3W_4W_7}(x,y,z)\mathrm{d}z\mathrm{d}y\mathrm{d}x} \\ &=& \frac{128/3e^4}{32/3e^4} \\ &=& 4\end{eqnarray*}$$ This answers your first question. For your second question, you're asked to compute the expected number of arrivals on $[0,7]$ given that the $5^\text{th}$ arrival occurs at time $t=4$. Since the number of arrivals on $(4,7]$ is $\text{Poisson}(2\cdot 3)$ we can say $$X_7|\{W_5=4\} \sim 5+P \text{ where } P \sim \text{Poisson}(6)$$ So $\mathbb{E}(X_7|W_5=4)=11$.