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This question asks about a variant of an alternating renewal process.

I am sitting at a cafe watching men and women walk by. The interarrival time $X$ between successive men is iid with distribution $F$, while the interarrival time $Y$ between successive women is iid with distribution $G$. Both $F$ and $G$ are nonlattice. Unlike the usual alternate renewal process, here we have two independent renewal processes, one for each sex.

Is there a nice way to compute the asymptotic probability that the most recent person seen is a man? (The usual theorem doesn't quite apply, since the interarrival times are within-gender only, not from person to person independent of gender.)

I'm interested both in general, and for the specific case that F and G are gamma distributions...

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the probability that the last person seen is a man is not clear for me. It's the probability that the most recent person is a man given... what? The past arrival times? Labeled? –  leonbloy Jul 3 '11 at 20:27
    
You say "we have two independent renewal processes". That is quite a restriction, and is not implied by your earlier words. –  Henry Jul 3 '11 at 20:48
    
Sorry if the statement wasn't clear. Let S_i be the arrival time of the i-th man, where S_i = S_{i-1} + X_i, and X_i is an independent variable with distribution F. (Take S_0 = 0.) Here F is a given distribution, such as a Gamma distribution with parameters k=2 and Theta=3. The arrival times for the women satisfy a similar distribution, with interarrival times Y_i satisfying another distribution G, such as a Gamma distribution with parameters k=4 and Theta=5. The men and women are arriving independently of each other; there is no interaction. –  Ron Rivest Jul 3 '11 at 21:11
    
Additional clarification: Let $P(t)$ be the event that at time $t$ the last person to walk by was a man. The quantity of interest is then $\lim_{t\rightarrow\infty} P(t)$, assuming this is defined. –  Ron Rivest Jul 3 '11 at 21:44
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2 Answers 2

First consider only one renewal process, describing i.i.d. interarrival times with integrable distribution $F$. Asymptotically, the distribution of the largest interval around the present time without any arrival is the size-biased transform of $F$ (more on this later) and the present time is uniformly distributed in this interval.

In other words, assume that $X$ is a random variable with distribution $F$. Then the size-biased transform of $F$ is the distribution of any random variable $\hat X$ such that $$ E(h(\hat X))=\frac{E(Xh(X))}{E(X)}, $$ for every bounded measurable function $h$. And the time elapsed since the last arrival is asymptotically distributed like $U\hat X$, where $U$ is uniform on $(0,1)$ and independent on $\hat X$.


Now, consider two independent renewal processes, with respective integrable distributions $F$ and $G$. The asymptotic probability that the last event corresponds to the $F$ renewal process is $$ p_F=P(U\hat X\le V\hat Y), $$ where $U$, $\hat X$, $V$ and $\hat Y$ are independent, $U$ and $V$ are uniform on $(0,1)$, the distribution of $\hat X$ is the size-biased transform of $F$ and the distribution of $\hat Y$ is the size-biased transform of $G$. Thus, $$ p_F=\frac{E(XY;UX<VY)}{E(X)E(Y)}, $$ where $X$, $Y$, $U$ and $V$ are independent, $U$ and $V$ are uniform on $(0,1)$, the distribution of $X$ is $F$ and the distribution of $Y$ is $G$.


Our next task is to get rid of $U$ and $V$. To condition on $(X,Y)$, one needs to compute $$ P(Ux<Vy)=[y<x]y/(2x)+[x<y](1-x/(2y)). $$ Unless I made a mistake, this yields

$$p_F=\frac{E(Y^2;Y<X)+E(2XY-X^2;X<Y)}{2E(X)E(Y)}.$$

Note that this is only one among several algebraically equivalent formulas for $p_F$.

All the expectations in the formula for $p_F$ are integrals involving $F$, $G$ and the respective densities $f$ and $g$. For example, $$ E(Y^2;Y<X)=E(Y^2(1-F(Y))=\int_0^{+\infty}y^2g(y)(1-F(y))\mathrm{d}y. $$ Post hoc checks: Here are some properties of the result above, which hold and ought to, for the formula to make sense.
(1) One has $p_F+p_G=1$ (where $p_G$ is the result one gets interchanging $X$ and $Y$) and $p_F$ is obviously positive, hence $p_G$ is positive as well. This proves that $p_F$ is in $(0,1)$.
(2) If $F=G$, everything cancels out in the numerator except the $E(2XY;X<Y)$ term, hence $p_F=\frac12$.
(3) If $X$ is exponential with parameter $a$ and $Y$ is exponential with parameter $b$, $p_F=a/(a+b)$.


Edit Regarding check (3) above, for exponential $a$ and $b$ distributions, one can compute everything in the formula giving $p_F$ as a function of $a$ and $b$, starting with $E(X)=1/a$, $E(Y)=1/b$, $f(x)=a\mathrm{e}^{-ax}$, $F(x)=1-\mathrm{e}^{-ax}$, $g(y)=b\mathrm{e}^{-by}$, $G(y)=1-\mathrm{e}^{-by}$.

Furthermore, $E(Y^2;Y<X)=E(Y^2(1-F(Y))$ hence $$ E(Y^2;Y<X)=\int_0^{+\infty}y^2b\mathrm{e}^{-by}\mathrm{e}^{-ay}\mathrm{d}y=\frac{2b}{(a+b)^3}, $$ by symmetry, $E(X^2;X<Y)=2a/(a+b)^3$, and finally, $$ E(XY;X<Y)=\int_0^{+\infty}ax\mathrm{e}^{-ax}\int_x^{+\infty}by\mathrm{e}^{-by}\mathrm{d}y\mathrm{d}x=\int_0^{+\infty}ax\mathrm{e}^{-ax}(x+1/b)\mathrm{e}^{-bx}\mathrm{d}x, $$ hence $E(XY;X<Y)=2a/(a+b)^3+a/(b(a+b)^2)$. Simplifying everything yields the value of $p_F=a/(a+b)$ given above.

Or, one can remember that for exponential interarrival times the arrival times form Poisson processes, that the superposition of two independent Poisson processes with intensities $a$ and $b$ is itself a Poisson process with intensity $a+b$, and finally that each point in the resulting point process is either an $a$-point or a $b$-point, independently of everything else and with respective probabilities $a/(a+b)$ and $b/(a+b)$. Putting all this together, one sees directly why the last event is an $a$-event with probability $p_F=a/(a+b)$.

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Excellent answer! This is very much what I am looking for! Thanks! Could you perhaps show a few details for the computation for (3) (two exponentials) if there is an easy way to see it is $a/(a+b)$?? –  Ron Rivest Jul 4 '11 at 17:44
    
Thanks for the appreciation. See edit. –  Did Jul 4 '11 at 19:48
    
Thanks again for the very clear explanation! Your final note about the superposition of two Poisson processes I was well aware of, but it is good to point out here; the details of the application of your earlier analysis was what I was asking for, and which you give. Thanks again! –  Ron Rivest Jul 5 '11 at 12:35
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If I understand right: arrivals of men are independent from arrivals of women. And each one is a renewal process, with distributions $f(t)$ and $g(t)$ (i.e. $f(t)$ is the probability density for the times between successive men arrivals $X$).

We take a random time $t$ (large, so that initial conditions do not matter and we can assume homogeneity) and call $A$ the "age" of the men arrival process, i.e., the time interval from the last men arrival to time $t$ ; analogously, we call $B$ the "age" of the women arrival. We want to compute the probability that the most recent is a man, that is $P(A < B)$

For this, we must compute the joint probability density of $A,B$. The marginal (density of the "age") is given by

$$p(A) = \frac{1}{E(X)} \int_A^\infty f(t) dt$$

(this can be easily found by a limit argument; check eg here, page 119).

And the same for $B$. And $A,B$ are indepent, hence the probality seeked is computed integrating in the region $A>B$

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