Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I would really appreciate if you could help out with the following question

Let $W_1,W_2,\ldots$ be the event times in a Poisson process $\{X(t);t\ge0\}$ of rate $\lambda$, and let $f(w)$ be an arbitrary function. Verify that $$ \mathbb{E}\left[\sum_{i=1}^{X(t)}f(W_i)\right]=\lambda\int_0^tf(w)\,dw. $$

(originally posted to this link: http://i.stack.imgur.com/3e6mf.png). I tried conditioning on X(t) = n but I am unsure how to proceed after that... Also I am not exactly sure how to show this is true since f(w) is an arbitrary function and I do not understand what that means.

Thanks a bunch for the help! I really

share|improve this question
    
Why don't you write out the question rather than just linking to it? –  George Lowther Mar 18 '11 at 1:01
    
I edited the question to add this. –  George Lowther Mar 18 '11 at 1:11
    
PASTA: "Poisson Arrivals See Time Averages" –  charles.y.zheng Mar 18 '11 at 1:34
    
@George i didn't know how to put math symbols on this forum, so I attached the image. but thanks for your response and thanks for transcribing it here. –  icobes Mar 18 '11 at 3:59
add comment

2 Answers

up vote 3 down vote accepted

Hint: Indeed, condition on $X(t)=n$. Then, note that given $X(t)=n$, the locations of the $n$ points are distributed as $n$ i.i.d. uniform$[0,t]$ rv's. This should lead you easily to the result, noting that ${\rm E}f(U)=\frac{1}{t}\int_0^t {f(w)dw} $, where $U$ is a uniform$[0,t]$ rv.

EDIT: More precisely (cf. cardinal's comment below), given $X(t)=n$, the points $W_1,\ldots,W_n$ are distributed as $n$ order statistics from the uniform distribution on $[0,t]$. However, the sum $f(W_1)+\cdots+f(W_n)$ (given $X(t)=n$) is equal in distribution to the sum $f(U_1)+\cdots+f(U_n)$ where the $U_i$ are i.i.d. uniform$[0,t]$ rv's, and thus the result follows straightforwardly using the hint.

EDIT (additional hint, in response to the OP's request): $$ \sum\nolimits_{n = 0}^\infty {n{\rm P}(X(t) = n)} = {\rm E}[X(t)] = ? $$ The left-hand side is given by $$ \sum\nolimits_{n = 0}^\infty {n{\rm E}[f(U)]{\rm P}(X(t) = n)}, $$ where $U$ is a uniform$[0,t]$ random variable.

EDIT (in response to the OP's request): In general, if $Y$ is a random variable with density function $h$, then ${\rm E}[f(Y)] = \int {f(y)h(y)dy} $. A uniform$[0,t]$ random variable, $U$, has constant density function $h(y)=1/t$ for $y \in [0,t]$ (and $0$ otherwise). From this it follows that ${\rm E}[f(U)] = \frac{1}{t}\int_0^t {f(y)dy} $.

As for the summation $\sum\nolimits_{n = 0}^\infty {n{\rm E}[f(U)]{\rm P}(X(t) = n)}$, first note that $$ {\rm E}\Big[\sum\nolimits_{i = 1}^{X(t)} {f(W_i )} \Big] = \sum\limits_{n = 0}^\infty {{\rm E}\Big[} \sum\nolimits_{i = 1}^n {f(W_i )} |X(t) = n \Big]{\rm P}(X(t) = n). $$ Now, recall that given $X(t)=n$, the sum $f(W_1)+\cdots+f(W_n)$ is equal in distribution to the sum $f(U_1)+\cdots+f(U_n)$, where the $U_i$ are i.i.d. uniform$[0,t]$ rv's. This accounts for the $n {\rm E}[f(U)]$ in the sum.

Further questions?

share|improve this answer
    
They are i.i.d. under a permutation of the indices of the $W_i$, but that's an admittedly minor point. –  cardinal Mar 17 '11 at 22:37
    
@cardinal: I'll edit the answer to make that point more precise. –  Shai Covo Mar 17 '11 at 22:41
    
Hi, I am stuck on this part: Sum from n = 0 to infinite of E[sum of i = 1 to n of f(Ui)] * P(X(t) = n). I find the expectation which is n/t and when multiplied by P(X(t) = n) that equals to 1/t * E[X(t)] but I am not sure what to do after... Thanks! –  icobes Mar 17 '11 at 23:38
    
@icobes: Note that $\sum\nolimits_{n = 0}^\infty {nP(X(t) = n)} = E[X(t)]$. Need more hints? –  Shai Covo Mar 17 '11 at 23:51
    
Yep, so I know that 1/t * E[X(t)] is the LHS. I also know that the expected value of a Poisson RV is lambda which makes it lambda * 1/t. But I have no clue how to make that equal to the RHS. A few more hints would be great! Thank you :) –  icobes Mar 17 '11 at 23:56
show 5 more comments

One way of solving this problem is to rewrite the sum as a Riemann-Stieltjes integral, $$ \sum_{i=1}^{X(t)}f(W_i)=\int_0^tf(s)\,dX_s. $$ Using the fact that $X(s)$ has expectation $\mathbb{E}[X(s)]=\lambda s$, you can take expectations* of this to get $$ \mathbb{E}\left[\sum_{i=1}^{X(t)}f(W_i)\right]=\lambda\int_0^tf(s)ds. $$

(*) To be rigorous, you should show that the expectation does indeed commute with the integral in this way. However, this is a general result which can be used in lots of situations, and is a very special case of the fact that stochastic integrals with respect to martingales are martingales ($X(s)-\lambda s$ is actually a martingale). In the case where $f$ is piecewise linear, the integral reduces to a sum, and expectations commute with summation by linearity. The fact that it extends to the general case of locally integrable $f$ is a consequence of the monotone class theorem.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.