# Independence of random sum variables

Let $(T_i)_{i \in \mathbb{N}}$ be a family of i.i.d. random variables where every $T_i \sim\mathrm{Exp}(\lambda)$. Now let $$Y :=\sum\limits_{j=1}^N T_j$$ such that for all $1 \leq j \leq N-1$ we have $T_j < c$, and $T_N \geq c$. In other words, we perform some experiment until it hits a certain threshold an sum up all outcomes. $N$ counts the number of trials we perform.

Is there a simple way to prove that $N$ is independent of all $T_j$ and also all $T_j \mid T_j < c$?

The context is an introductory course for computer science undergrads; they don't really know about $\sigma$-algebras or any measure theory, just in case that becomes relevant here.

$N$ is certainly not independent of any $T_j$. That is the conditional probability $P(T_j \ge c\mid N = j) = 1$ while $P(T_j \ge c) < 1$. If $N$ and $T_j$ were independent these would be equal.

• Thank you, is there anything to be said about a possible independence between $N$ and the conditional variables $T_j | T_j < c$? Jun 17, 2014 at 21:33

Your notation is hopelessly ambiguous: "$T\sim\mathrm{Exp}(\lambda)$" sometimes means $P(T>t)=e^{-\lambda t}$ (so that $E(T)=1/\lambda$) and sometimes means $P(T>t)=e^{-t/\lambda}$ (so that $E(T)=\lambda$). I'll just let $E(T)=\mu$, so that $P(T>t)=e^{-t/\mu}$.

The probability distribution of $N-1$ is Poisson with expected value $c/\mu$. When the waiting time until the next occurence is exponential, then the number of occurrences within a specified time is Poisson-distributed.

At any rate, $N$ is certainly not independent of $T_j$ for any value of $j$.

• Thanks, I was not aware that $T$ ~ Exp($\lambda$) was ambiguous. Jun 17, 2014 at 21:34
• I think the question is incorrectly stated. Note that $Y$ was not mentioned after its definition. Did you mean to define: $$Y_n=\sum_{j=1}^n T_j$$ and then define $N=n \iff Y_{n-1}<c$ and $Y_n \ge c.$ Then maybe it was a question on whether $N$ was a stopping time? Jun 22, 2014 at 7:39