Expectation for a Poisson processes

Question

This is a very elementary question I am asking where I want to make sure my procedure is right. I am asked to calculate: $V=E[X_2\mid N(1)=1,N(2)=2,N(3)=3]$, where $N(t)$ is the counting function of the Poisson process that has intensity $\lambda$, and $X_2$ is the total holding time at count $n=2$. Does it makes sense to decompose the Expectation like this:

$$V=E[X_2\mid N(1)=1]P(N(1)=1)+E[X_2\mid N(2)=2]P(N(2)=2)+E[X_2\mid N(3)=3]P(N(3)=3).$$

I'm assuming the justification from above is something like the law of total expectation?

Answer 1

This is a nice question... Here are two approaches, the first one entirely basic, the second one more elaborate and (as was to be expected) considerably shorter.

Hands-on approach: The event $A=[N_1=1,N_2=2,N_3=3]$ is $$[X_0\lt1\lt X_0+X_1\lt 2\lt X_0+X_1+X_2\lt3\lt X_0+X_1+X_2+X_3], $$ where, for every $n\geqslant0$, $X_n$ denotes the holding time at count $n$. The random variables $(X_n)_{0\leqslant n\leqslant 3}$ are i.i.d. and exponential with parameter $\lambda$ hence $P[X_3\geqslant x]=\mathrm e^{-\lambda x}$ for every $x\geqslant0$ and, for every nonnegative parameter $i$, $$ E[X_2^i;A]=\mathrm e^{-3\lambda}E[X_2^i\mathrm e^{\lambda X_0}\mathrm e^{\lambda X_1}\mathrm e^{\lambda X_2};B], $$ where $$ B=[X_0\lt1\lt X_0+X_1\lt 2\lt X_0+X_1+X_2\lt3]. $$ The exponentials in the last expectation cancel almost perfectly the densities hence $$ E[X_2^i;A]=\lambda^3\mathrm e^{-3\lambda}C_i,\qquad C_i=\int_{\Delta}x_2^i\mathrm dx_0\mathrm dx_1\mathrm dx_2, $$ where $$ \Delta=\{(x_0,x_1,x_2)\mid 0\lt x_0\lt1\lt x_0+x_1\lt 2\lt x_0+x_1+x_2\lt3\}. $$ The transformation $(x_0,x_1,x_2)\mapsto(y_0,y_1,y_2)=(x_0,x_0+x_1,x_0+x_1+x_2)$ preserves the volumes and sends $\Delta$ to the cube $D=(0,1)\times(1,2)\times(2,3)$. Thus, $C_0=|\Delta|$ is $C_0=|D|=1$ and, using the fact that $x_2=y_2-y_1$, $$ C_1=\int_D(y_2-y_1)\mathrm dy_0\mathrm dy_1\mathrm dy_2=\int_2^3y_2\mathrm dy_2-\int_1^2y_1\mathrm dy_1=\tfrac52-\tfrac32=1. $$ Finally, $$ E[X_2\mid A]=1. $$ More elaborate approach: Conditionally on the event $A$, the time $L_2$ of the last increase of the Poisson process before time $2$ and the time $R_2$ of its first increase after time $2$ are (independent and) uniform on the intervals $(1,2)$ and $(2,3)$ respectively. Since $X_2=R_2-L_2$, $E[X_2\mid A]=E[U_2-U_1]$ where each $U_n$ is uniform on the interval $(n,n+1)$. Hence $E[U_n]=n+\frac12$ for every $n$, which implies $$E[X_2\mid A]=1.$$

Answer 2

If $N(1) = 1, N(2) = 2$ and $N(3) = 3$, then you know that the first arrival occurred at or before $t = 1$ and that the second arrival occurred between $t = 1$ and $t = 2$. Since arrival times are exponential random variables (which are also memoryless random variables), the conditional distribution of $X_2-1$ given that $X_2 > 1$ (how can we be sure of that?) is an exponential random variable. But you also know that $X_2 \leq 2$, and so the conditional distribution of $X_2-1$ given that $N(1) = 1, N(2) = 2$ (and $N(3) = 3$) is a truncated exponential distribution. So find the density and compute the conditional mean of $X_2$.