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 $\mathbb{E}(Y_m)=m$.

Thanks in advance for your time and help!   
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?
 A: 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 


*

*Property (P1): parameter $λ(m+1-m)=1$ (since $λ=1$ is given), and

*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 $$S_k\le m<S_k+X_{k+1}$$ 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 m<S_k+X_{k+1})
\\[0.2cm]&=P(X_{k+1}+\dots+X_{k+n}>m+1-S_k \mid 0\le m-S_k<X_{k+1})\\[0.2cm]&=P(X_{k+1}+\dots+X_{k+n}>1)\end{align} which is independent of $F_m$ due to the independence of the $X_i$'s for all $i \in \mathbb N$. 
