2
$\begingroup$

Assume $X_1,\ldots,X_n$ are i.i.d exponential random variables with pdf $\lambda e^{\lambda x}$, and $Y_1,\ldots,Y_n$ are i.i.d exponential random variables, independent of $X_i$s, and with pdf $\lambda' e^{\lambda' x}$, where $\lambda'<\lambda$. Calculate $$p_n=\mathbb{P}\left(X_1<Y_1 \bigcap X_1+X_2<Y_1+Y_2 \bigcap \ldots \bigcap \sum_{i=1}^n X_i \leq \sum_{i=1}^n Y_i\right)$$ see if $\lim\limits_{n \to \infty} p_n =c>0$.

I can say: $$p_n = \mathbb{P}\left(X_1<Y_1\right) \mathbb{P}\left(X_1+X_2<Y_1+Y_2|X_1<Y_1\right) \ldots \mathbb{P}\left( \sum_{i=1}^n X_i \leq \sum_{i=1}^n Y_i\bigg|X_1<Y_1 \bigcap X_1+X_2<Y_1+Y_2 \bigcap \ldots \right)$$

But, I am not sure how to continue. Also, if I want to just consider the asymptotic behavior, I can use bounds: $$p_n\geq \mathbb{P}\left(X_1<Y_1 \bigcap X_2<Y_2 \bigcap \ldots \bigcap X_n<Y_n\right) = \mathbb{P}\left(X_1<Y_1\right)^n \to 0$$

which did not help. Any idea?

$\endgroup$
4
  • 1
    $\begingroup$ It looks like a generalization of the Bertrand's ballot problem (kurser.math.su.se/pluginfile.php/16103/mod_folder/content/0/…) $\endgroup$
    – Jean Marie
    May 22, 2017 at 20:23
  • $\begingroup$ From the problem statement you don't need to compute the probability, rather it seems, you need to bound it. Have you tried anything along these lines? $\endgroup$
    – David
    May 22, 2017 at 21:23
  • $\begingroup$ @david yes, but it didnt work since i achieved p>=0. $\endgroup$ May 22, 2017 at 21:24
  • $\begingroup$ @david Hi, I just updated the statement to show what I tried for bounds. $\endgroup$ May 23, 2017 at 0:42

1 Answer 1

1
$\begingroup$

You can think of $\sum_{i=1}^j X_i$ as the $j$th arrival of a Poisson process with rate $\lambda$; similarly for $\sum_{i=1}^j Y_i$ and a Poisson process with rate $\lambda'$.

Since these processes are independence, superposition implies that you can think of this as a combined Poisson process with rate $\lambda + \lambda'$, and then for each arrival, flip a coin with heads probability $q:=\frac{\lambda}{\lambda+\lambda'}$; if it is heads, label the arrival "heads," and otherwise label the arrival as "tails." The heads arrivals form the first Poisson process with rate $\lambda$, and the tails arrivals form the Poisson process with rate $\lambda'$.

Thus if you flip a coin with heads probability $q$ repeatedly, then $p_n$ is the probability that the $j$th heads is before the $j$th tails flip, for each $j=1,\ldots,n$.

However, I am not sure how to compute this. As noted in the comments, this is similar to Bertrand's ballot problem; see also this post.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .