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Let $W_1, W_2, \dots $ be i.i.d. exponential $(\lambda )$ variables. For each $n$, find the distribution of $T_n = W_1 + W_2 + \cdots + W_n$

I know how to deal with two i.i.d random exponential variables. Namely, I know that for two independent exponential random variables $X$ and $Y$ and $S$ = $X+Y$, I can use the discrete convolution formula to write:

$P(S=s) = \sum_{all x} P(X=x, Y=s-x)$. My problem is I do not know how to expand this to $n$ variables to answer my problem

Edit: I think this could be an induction problem:

For $T_1$ the distribution is just $W_1$, or the exponential distribution function.

For $T_2$ the distribution is $T_1 + W_2$, which would be the addition of two exponential distribution functions.

For $T_n$ then, the distribution is $T_{n-1} + W_n$ or the sume of $n$ exponential distribution functions.

Does this make sense and is this correct? How would I correctly set up the induction?

Jayant.M
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1 Answers1

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We can solve a more general case, consider the sum of two gamma r.v.'s $X$ and $Y$ with PDF: $$X \sim \frac {b^ax^{a-1}e^{-bx}}{\Gamma (a)} \quad Y\sim \frac {b^ty^{t-1}e^{-by}}{\Gamma (t)}$$ Now $$\begin{align} f_{X+Y}(w) &=C\int_0^{w} y^{t-1}e^{-by}(w-y)^{a-1}e^{-b(w-y)}\mathrm{d}y \\ &=Ce^{-bw}\int_0^{w} y^{t-1}(w-y)^{a-1}\mathrm{d}y \\ &= Ce^{-bw}w^{a+t-1}\int_0^{1} y^{t-1}(1-u)^{a-1}\mathrm{d}y \quad \text{setting} \ u = \frac yw \\ &=De^{-bw}w^{a+t-1} \quad \text{evaluating the integral to a constant and combining with C} \end{align} $$ Since $f$ is a density function, $\int f=1$, i.e. $\int_0^{\infty} De^{-bw}w^{a+t-1} \mathrm d a=1$ meaning that $$D = \frac {1}{\int_0^{\infty}e^{-bw}w^{a+t-1} \mathrm d a}=\frac {1}{\Gamma (a+t)}$$ Therefore, $$f_{X+Y}(w)=\frac{e^{-bw}w^{a+t-1}}{\Gamma (a+t)}$$ or $X+Y \sim \text{Gamma} (a+t,b)$ By induction, if $X_1,...,X_n$ are independent and $X_i \sim \text{Gamma} (a_i, b)$ then $\sum X_i \sim \text{Gamma} \left (\sum a_i, b\right)$. Now notice that an exponential random variable is a gamma random variable with $a=1$.

GuPe
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