Proof that process is martingale, exponential distribution

Let $X_1,X_2,\dots$ be i.i.d. random variables with exponential distribution with parameter $1$ and define $$Y_m= \sup{\{k\ge1:X_1+\dots+X_k\le m\}}$$ Prove that $Y_m-m$ is martingale and $\mathbb{E}(Y_m)=m$.

My attempts: We want to prove that $\mathbb E[Y_{m+1}-Y_m\mid F_m]=1$, but I don't know how to connect values of $Y_m$ and $Y_{m+1}$.
Edit
$Y_m$ has the Poisson distribution with parameter m. It's enough to prove that $Y_{m+1}-Y_m$ is independent of $F_m$. Could anyone help me with this?

• If we take $m\ge 0$ to be a continuous variable, then $\{Y_m: m\ge 0\}$ is a unit rate Poisson process. Commented Jan 24, 2016 at 16:49
• Oh, I don't know what Poisson process is. I will read about it. How does it helps? Commented Jan 24, 2016 at 16:58
• It lets you know that $Y_m$ has stationary independent increments, which makes the martingale calculation easy. Commented Jan 24, 2016 at 17:13
• Could you recommend good book/pdf about it available online? Commented Jan 24, 2016 at 17:59

The process $$\{Y_m\}_{m\ge 1}$$ is a Poisson process that counts the events up to time $$m \in \Bbb R_+$$. The expression $$\tilde Y_1:=Y_{m+1}-Y_m$$ counts the number of events in the time period $$(m, m+1]$$ and due to the memoryless property of the exponential distribution it can be proven that it has again the Poisson distribution with

1. Property (P1): parameter $$λ(m+1-m)=1$$ (since $$λ=1$$ is given), and
2. Property (P2): is independent from the $$\{Y_t\}_{0\le t\le m}$$.

Hence, since $$Y_m$$ is known given $$F_m$$ we can write \begin{align}\Bbb E[Y_{m+1}-(m+1)\mid F_m]&=\Bbb E[Y_{m+1}\pm Y_m-(m+1)\mid F_m]=\\[0.2cm]&=\Bbb E[Y_{m+1}-Y_m\mid F_m]+\Bbb E[Y_m-(m+1)\mid F_m]=^{(P2)}\\[0.2cm]&=\Bbb E[Y_{m+1}-Y_m]+Y_m-(m+1)=^{(P1)}\\[0.2cm]&=1+Y_m-m-1=Y_m-m\end{align} which proves (together with $$E[|Y_m|]<+\infty$$) that the process $$\{Y_m\}_{m\ge 1}$$ is a martingale.

Edit: To prove (P2), i.e. that $$Y_{m+1}-Y_m$$ is independent of $$F_m$$, note that given $$F_m$$ we know the value of $$Y_m$$, say $$Y_m=k$$ and this is equivalent to $$S_k\le m Now for $$n\in \mathbb N$$ we want to show that the event $$Y_{m+1}-Y_m< n$$ is independent from $$F_m$$. Given the value of $$Y_m=k$$ we can write the probability of this event as: \begin{align}P(Y_{m+1}-Y_m< n \mid F_m) &= P(Y_{m+1}< k+n \mid Y_m=k)\\[0.2cm]&=P(S_{k+n}>m+1 \mid S_k\le mm+1-S_k \mid 0\le m-S_k1)\end{align} which is independent of $$F_m$$ due to the independence of the $$X_i$$'s for all $$i \in \mathbb N$$.

• Thank you very much. Could you explain the last equality? Commented Jan 25, 2016 at 19:17
• @Cofibration There was a typographical mistake in the last equality. I corrected it. Practically it says that if you know that $S_k<m$ (given $F_m$) then $S_{k+n}=S_k+X_{k+1}+\ldots+X_{k+n}$ where $S_k$ is known and $X_{k+1},\dots, X_{k+n}$ are independent from what happened previously. Commented Jan 25, 2016 at 19:22
• @Cofibration I made an edit. I hope that it is more clear as an explanation now. Commented Jan 26, 2016 at 9:23
• It helped a lot. :) Thank you! Commented Jan 26, 2016 at 11:10