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How to prove using induction the following is true : Let $X_1, X_2,\ldots,X_n$ be i.i.d r.v. that are distributed exponentially with $\lambda$. Then $ \forall n \ge 3$ $$ \Pr(X_1 > \sum_{k=2}^n X_k) = \frac 1 {2^{n-1}} $$

I was able to show the basis step, where $n=3$ $$ \Pr(X_1 > X_2+X_3) = \int_0^\infty \Pr(X_1 > x + y) f(x) f(y) \, dx\, dy = \int_0^\infty e^{-\lambda (x+y)} \frac 1 \lambda e^{-\lambda x} e^{-\lambda y} \,dx\, dy = \left.- \frac 1 2 e^{-2\lambda x}\right]_0^\infty \cdot \left.\frac {- 1} 2 e^{-2\lambda x}\right]_0^\infty = \frac 1 {2^{3-1}} $$

Now am I supposed to go about solving for the inductive step, it says in the question to use the memoryless property when proving for this.

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Use the fact that for exponential random variables $X$, $$\Pr[X>s+t|X>s] = \Pr[X>t].$$ Using this, we get $$ \Pr[X_1>X_2+\cdots+X_n] = \\ \Pr[X_1>X_2+\cdots+X_n|X_1>X_2+\cdots+x_{n-1}] \Pr[X_1>X_2+\cdots+X_{n-1}] = \\ \Pr[X_1 > X_n] \Pr[X_1>X_2+\cdots+X_{n-1}]. $$ Now you can use induction. What we use here is that if $X_1>X_2+\cdots+X_n$ then also $X_1>X_2+\cdots+X_{n-1}$, since $X_n \geq 0$.

Even without (formally) using induction, in this way you easily get $$ \Pr[X_1>X_2+\cdots+X_n] = \Pr[X_1>X_2] \cdots \Pr[X_1>X_n] = \frac{1}{2^{n-1}}, $$ due to symmetry and the fact that exponential random variables have no atoms (so $\Pr[X_1 = X_2] = 0$).

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