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Given a poisson process $P_t$ with rate $r$, with arrival times $S_n$

How do I calculate the Variance of $S_2-S_1|P_t=2$?

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What do you call $P_t$, if not $S_t$ (and vice versa)? – Did Apr 6 '13 at 10:44
So, $[P_t=2]=[S_2\leqslant t\lt S_3]$? – Did Apr 6 '13 at 11:56
You probably mean $S_1\leqslant S_2\leqslant t$. See answer below. – Did Apr 6 '13 at 12:58

1. Conditionally on $S_2=x$ with $0\lt x\lt t$ and on $P_t=2$, $S_1$ is uniformly distributed on $(0,x)$, hence $$ E[S_2-S_1\mid P_t=2]=\tfrac12E[S_2\mid P_t=2], $$ and $$ E[(S_2-S_1)^2\mid P_t=2]=\tfrac13E[S_2^2\mid P_t=2]. $$ 2. The density $f$ of the (absolute) distribution of $S_2$ is such that $f(x)=r^2x\mathrm e^{-rx}$ for every $x\geqslant0$. Furthermore, $P_t=2$ means that $S_2\leqslant t$ and $S_3-S_2\geqslant t-S_2$. Hence, for every bounded function $u$, $$ E[u(S_2);P_t=2]=\int_0^t u(x)f(x)\mathrm e^{-r(t-x)}\mathrm dx=r^2\mathrm e^{-rt}\int_0^t u(x)x\mathrm dx. $$ Thus, the density $g_t$ of the distribution of $S_2$ conditionally on $P_t=2$ is such that $g_t(x)$ is proportional to $x$ on $0\leqslant x\leqslant t$, that is, $g_t(x)=2t^{-2}x$ on $0\leqslant x\leqslant t$. In particular, for every $n$, $$ E[S_2^n\mid P_t=2]=\int_0^tx^ng_t(x)\mathrm dx=\tfrac2{n+2}t^n. $$ 3. The first computation above yields $$ \mathrm{var}(S_2-S_1\mid P_t=2)=\tfrac13E[S_2^2\mid P_t=2]-\tfrac14E[S_2\mid P_t=2]^2, $$ and the second computation yields $$ \mathrm{var}(S_2-S_1\mid P_t=2)=\tfrac13\tfrac24t^2-\tfrac14\left(\tfrac23t\right)^2=\tfrac1{18}t^2. $$

Edit: According to the comments, the OP also wants to compute the CDF of $(S_1,S_2)$ conditionally on $[P_t=2]$. This is not the most direct route to solve the question asked in the post but here is a way to do that. Let us consider $G(x,y)=P[S_1\leqslant x,S_2\leqslant y,P_t=2]$. Since $S_1$ is exponentially distributed, this yields $$ G(x,y)=\int_0^xH(u,y)\,r\mathrm e^{-ru}\mathrm du,\qquad H(x,y)= P[S_2\leqslant y,P_t=2\mid S_1=u]. $$ By the independence property of the Poisson process, $$ H(u,y)=P[S_1\leqslant y-u,P_{t-u}=1]=\int_0^{y-u}\,r\mathrm e^{-rv}P[P_{t-u-v}=0]\mathrm dv, $$ hence, $$ H(u,y)=\int_0^{y-u}\,r\mathrm e^{-rv}\mathrm e^{-r(t-u-v)}\mathrm dv. $$ In terms of $G(x,y)$, this yields $$ G(x,y)=\int_0^x\int_0^{y-u}\,r^2\mathrm e^{-rt}\mathrm dv\mathrm du=\tfrac12r^2\mathrm e^{-rt}x(2y-x). $$ Finally, $P[S_1\leqslant x,S_2\leqslant y\mid P_t=2]=G(x,y)/G(t,t)$ hence, for every $0\leqslant x\leqslant y\leqslant t$, $$ P[S_1\leqslant x,S_2\leqslant y\mid P_t=2]=t^{-2}x(2y-x). $$ Note for example that $[S_2\leqslant y]=[S_1\leqslant y,S_2\leqslant y]$ hence $$ P[S_2\leqslant y\mid P_t=2]=t^{-2}y^2, $$ which yields the density $g_t$ computed above. Note finally that the formula for the CDF of $(S_1,S_2)$ conditionally on $[P_t=2]$ given above is the CDF of the uniform density on the triangle $0\leqslant x\leqslant y\leqslant t$, as the following property of homogenous Poisson processes confirms:

For every $n\geqslant1$, conditionally on $[P_t=n]$, the set $\{S_k\mid 1\leqslant k\leqslant n\}$ is distributed like the set $\{X_k \mid 1\leqslant k\leqslant n\}$, where $(X_k)_{1\leqslant k\leqslant n}$ is uniform on $[0,t]^n$, or, equivalently, is i.i.d. uniform on $[0,t]$. Thus, $(S_k)_{1\leqslant k\leqslant n}$ is distributed like the ordered sample $(X_{(k)})_{1\leqslant k\leqslant n}$.

If one is "allowed" to use this description, things become simpler since all of the above can be replaced by the identity $$ \mathrm{var}(S_2-S_1\mid P_t=2)=E[(X_2-X_1)^2]-E[X_{(2)}-X_{(1)}]^2=2\mathrm{var}(X)-E[X_{(2)}-X_{(1)}]^2, $$ where $X$ is uniform on $[0,t]$. One can guess that, by symmetry, $E[X_{(2)}-X_{(1)}]=\frac13t$. Furthermore, $E[X]=\frac12t$ and $E[X^2]=\frac13t^2$, hence $\mathrm{var}(X)=\frac13t^2-\left(\frac12t\right)^2=\frac1{12}t^2$. And once again, the final result $\mathrm{var}(S_2-S_1\mid P_t=2)=2\frac1{12}t^2-\left(\frac13t\right)^2=\tfrac1{18}t^2. $

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Your RHS does not depend on $s_2$ (even assuming $t_1$ means $s_1$). It should. – Did Apr 6 '13 at 21:37
At this point, I think it would be best if you showed how you reached this formula, otherwise I am afraid we will be running in circles. – Did Apr 6 '13 at 22:11
The complement of $[P_{s_1}\geqslant1,P_{s_2}\geqslant2]$ is not $[P_{s_1}\lt1,P_{s_2}\lt2]$. – Did Apr 7 '13 at 0:26
Not sure this new attempt is fully correct, please compare with the Edit. – Did Apr 7 '13 at 9:46
Yes, if what you mean is that $S_1$ conditionally on $[S_2=y,P_t=2]$ is uniformly distributed on $(0,y)$. – Did Apr 7 '13 at 11:55

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