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We are given independent identically distributed random variables $X_1, X_2, \ldots \sim \text{Exponential}(1),$ and we define $$Y = {\sum_{i=1}^\infty {2^{-i}} \sum_{j=1}^i {X_j}}.$$

The question asks about finding the distribution of $Y.$

We can simplify the RHS as, $$Y = 2 {\sum_{j=1}^{\infty} {2^{-j} X_j}} . $$

And after that so far, I have tried computing the MGF of $Y$ as, $$\prod_{k=0}^{\infty} \frac{1}{1 - \frac{t}{2^k}}$$ and the inner expression can be rewritten as $$\frac{1}{1 - \frac{t}{2^k}} = \sum_{j=0}^{\infty} \left( \frac{t}{2^k} \right)^j.$$

Not sure how to proceed after this. I think that to define a distribution, we need to compute the support of the random variable. Meaning we need the CDF: $P(Y \leq y) $?

pinton jol
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1 Answers1

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The PDF can be written in terms of an infinite series, by performing a partial fraction decomposition on the Laplace transform. First, generalize slightly by assuming $X_k\sim \text{Exp}(\lambda)$ and $Y_N=\sum_{k=1}^N\frac{a^k}{1-a}X_k$. In this notation, $Y=\lim_{N\to \infty} Y_N$. Then

$$\mathbb{E}(e^{-sY_N})=\prod_{k=1}^N\frac{a^{-k}}{a^{-k}+\frac{s}{\lambda(1-a)}}$$

Define $G_{k}(a)=\prod_{m=1}^k(1-a^m)$. Performing the partial fraction decomposition yields

$$\mathbb{E}(e^{-sY_N})=\sum_{k=1}^N\frac{1}{\frac{s}{\lambda(1-a)}+a^{-k}}\frac{a^{-k}}{G_{k-1}(1/a)G_{N-k}(a)}$$

We can now invert the LT and, using a generalization of Tannery's theorem we can show that

$$f_Y(y)=\lambda(1-a)G^{-1}_{\infty}(a)\sum_{k=1}^\infty\frac{e^{-\frac{\lambda (1-a)}{a^k}y}}{a^kG_{k-1}(1/a)}$$

The series converges rapidly, as a double exponential. This form is not particularly illuminating or amenable to further simplification; it is possible, however, to analytically compute the cumulants of $Y$. In fact, one can show that

$$\kappa_n=\frac{(n-1)!}{1-a^n}\left(\frac{a}{\lambda(1-a)}\right)^n$$

K. Grammatikos
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