How to prove $~P(X_1+\cdots+X_nQuestion : How to prove below expression?
$$~P(X_1+\cdots+X_n<Y)=P(X_1<Y)^n~$$
We have memoryless property
$$P(Y>X_1+X_2|Y>X_1)=P(Y>X_2)$$
and $X_1,\cdots,X_n$ be independent and nonnegative random variables. A random variable $~Y~$ is exponentially distributed.
I tried to divide $~P(X_1+\cdots+X_n<Y)~$ by $~P(X_1<Y)~$ for $~n~$ times, but no effect.
and tried again to divide $~P(X_1+\cdots+X_n<Y)~$  by $~P(X_n<Y), \cdots, P(X_2<Y)~$, same happened again. 
 A: Consider Memoryless property
\begin{equation}
\begin{split}
P(Y > X_1+X_2/Y>X_1) &= P(Y>X_2) \\
\frac{P(Y > X_1+X_2, Y > X_1)}{P(Y > X_1)} &= P(Y>X_2) \\
\frac{P(Y > X_1+X_2)}{P(Y > X_1)} &= P(Y>X_2) \\
P(Y > X_1+X_2) &= P(Y>X_1)P(y > X_2)
\end{split}
\end{equation}
Similarly 
\begin{equation}
\begin{split}
P(Y > X_1+(X_2+X_3)/Y>X_1) &= P(Y>(X_2+X_3)) \\
P(Y > X_1+X_2+X_3) &= P(Y>X_1)P(y > X_2+X_3) \\
P(Y > X_1+X_2+X_3) &= P(Y>X_1)P(y > X_2) P(Y > X_3)
\end{split}
\end{equation}
You can extend this for $n$ variables in the same way. If all $X$ s are iid, then one can write
$P(Y > X_1+X_2+..+X_n) = P(Y>X_1)^n$
A: You can use an inductive proof here.  The base case where $n=1$ is a trivial tautology, so it remains only to establish the inductive step.  That is, we let $S_n \equiv X_1+\cdots+X_n$ and we assume that:
$$\mathbb{P}(S_n<Y) = \mathbb{P}(X_1<Y)^n.$$
Using the memoryless property and the inductive assumption we then have:
$$\begin{equation} \begin{aligned}
\mathbb{P}(S_{n+1}<Y) 
&= \mathbb{P}(S_n+X_{n+1}<Y ) \\[6pt]
&= \mathbb{P}(S_n+X_{n+1}<Y | S_n < Y ) \cdot \mathbb{P}(S_n < Y) \\[6pt]
&= \mathbb{P}(X_{n+1}<Y ) \cdot \mathbb{P}(S_n < Y) \\[6pt]
&= \mathbb{P}(X_{n+1}<Y ) \cdot \mathbb{P}(X_1<Y)^n \\[6pt]
&= \mathbb{P}(X_1<Y)^{n+1}. \\[6pt]
\end{aligned} \end{equation}$$
(The last step follows from the fact that the random variables $X_1,X_2,X_3,...$ are identically distributed.)
