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Let's say $X_1,X_2,...,X_n$ are a sequence of i.i.d random variables from some unknown distribution. If we know that summation of them, $Y$ = $\sum_{i=1}^{n} X_i$, has exponential distribution with parameter $\lambda$, what is distribution of $X_i$?

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  • $\begingroup$ Do you know the relationship between $\lambda$ ans $n$? Is this property valid for all $n$ or for one specific fixed one? $\endgroup$ – gt6989b Feb 22 '18 at 23:48
  • $\begingroup$ There is no specific relation between $n$, $\lambda$, but we can assume that we know value of $\lambda$. $n$ could be any arbitrary number.I can find m.g.f of $X_i$. Since it is: $g(X_i)$ = $\sqrt[n]{g(expnential)}$ $\endgroup$ – Yashar Feb 22 '18 at 23:56
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Existence. A gamma distribution, denoted by $\operatorname{Gamma}(k, \lambda)$ for parameters $k, \lambda > 0$, is defined as a distribution supported on $[0,\infty)$ with the p.d.f.

$$ f(x) = \frac{\lambda^k x^{k-1} e^{-\lambda x}}{\Gamma(k)} \mathbf{1}_{(0,\infty)}(x). $$

Now if $X \sim \operatorname{Gamma}(k,\lambda)$ and $Y \sim \operatorname{Gamma}(l,\lambda)$ are independent, then $Z = X+Y$ has density $f_Z$ which satisfies

\begin{align*} \forall x > 0 \ : \quad f_Z(x) &= \int_{0}^{x} f_X(t)f_Y(x-t) \, dt \\ &= \int_{0}^{x} \frac{\lambda^k t^{k-1} e^{-\lambda t}}{\Gamma(k)} \cdot \frac{\lambda^l (x-t)^{l-1} e^{-\lambda (x-t)}}{\Gamma(l)} \, dt \\ &= \frac{\lambda^{k+l}e^{-\lambda x}}{\Gamma(k)\Gamma(l)} \int_{0}^{x} t^{k-1}(x-t)^{l-1} \, dt \\ &= \frac{\lambda^{k+l}x^{k+l-1}e^{-\lambda x}}{\Gamma(k+l)} \end{align*}

by the beta function identity. So $Z \sim \operatorname{Gamma}(k+l,\lambda)$. An immediate corollary is

Proposition. If $X_1, \cdots, X_n$ are independent and $X_i \sim \operatorname{Gamma}(k_i, \lambda)$ with $k_i, \lambda > 0$ for each $i$, then $\sum_{i=1}^{n} X_i \sim \operatorname{Gamma}(\sum_{i=1}^{n}k_i, \lambda)$.

Now by noting that $\operatorname{Exp}(\lambda) = \operatorname{Gamma}(1,\lambda)$, we find that $\operatorname{Gamma}(\frac{1}{n}, \lambda)$ is one possible choice.

Uniqueness. Notice that $Y$ has the m.g.f.

$$ \mathsf{E}[e^{sY}] = \frac{\lambda}{\lambda - s} $$

for $s < \lambda$. So it follows that

$$ \mathsf{E}[e^{sX_1}] = \left( \frac{\lambda}{\lambda - s} \right)^{1/n}. $$

Since the m.g.f. determines a distribution as long as it exists on a neighborhood of $0$, there is at most one choice of the distribution of $X_1$.

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