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I have $N$ Independent Random variables Laplacian distributions with $\mu=0$ and positive $b=\sigma^2/2$. I also have dominant random variable $(X_s)$ with Laplacian distribution with $\mu=0$ and $b=\sigma^2_s/2.$ $(\sigma^2_s \gg \sigma^2).$ I want to find the joint distribution for $$X= \sum_1^N X_i +X_s$$

My work: I took the product of the CF of the random variables on the right to get

$$ \varphi_X(t)= \frac{2}{2+\sigma^2_s t^2} \left( \frac{2}{2+\sigma^2t^2} \right)^N$$

The denominator of this expression can be simplified to

\begin{align} & (2+\sigma^2_s t^2) (2+\sigma^2t^2)^N \\[6pt] = {} & (2+\sigma^2_s t^2)(\sigma^2_s)^N(2/\sigma^2_s + \sigma^2/\sigma^2_s t^2)^N \\[6pt] \approx {} & (2+\sigma_s^2 t^2)2^N \end{align}

Thus the CF will be that of the $X_s$.

I would like to get some approximate form of the distribution of $X$ based on $N.$ In this approach, I am not able to. Can someone please help?

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  • $\begingroup$ Related: mathoverflow.net/questions/66763/… $\endgroup$
    – Galen
    Commented Dec 6, 2023 at 15:44
  • $\begingroup$ Since they are independent, you can convolve their distributions. Convolution theorem might be handy. $\endgroup$
    – Galen
    Commented Dec 6, 2023 at 15:47
  • $\begingroup$ One approach is to express each $X_i$ as a mixture of $Y_i$ and $-Z_i$ where $Y_i$ and $Z_i$ are independent Gamma$(1,\sigma)$ variable and use the result at stats.stackexchange.com/questions/72479 (which also includes a link to a paper on approximating such sums when $N$ is large). // I don't follow your "thus:" for large $N$ that's a very poor approximation. You need to consider higher powers of $t,$ at least when $N\sigma^2/\sigma_s^2\gg 0.$ // What is "$\sigma_n^2$"? A typo? $\endgroup$
    – whuber ♦
    Commented Dec 6, 2023 at 16:06
  • $\begingroup$ @Galen Then I would have to convolve these N+1 Laplacian distributions right? That will be a complex distribution I think. $\endgroup$
    – lone_wolf
    Commented Dec 6, 2023 at 16:18
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    $\begingroup$ To point you in the correct direction, I wish to remark that a correct analysis of the cf can be obtained via the Binomial Theorem as $$\phi_X(t)=(1+\sigma_s^2t^2/2)^{-1}(1+\sigma^2t^2/2)^{-N}=\sum_{j=0}^\infty\binom{-1}{j}(-\sigma_s^2t^2/2)^j\sum_{k=0}^\infty\binom{-N}{k}(-\sigma^2t^2/2)^k.$$ Expanding in powers of $t$ gives a series $$1+(\sigma_s^2+N\sigma^2)\left(\frac{-t^2}{2!}\right)+3\left(\sigma_s^4+\frac{N(N+1)}{2}\sigma^4+N\sigma_s^2\sigma^2\right)\left(\frac{t^4}{4!}\right)+\cdots.$$ $N$ shows up explicitly, as we would hope. $\endgroup$
    – whuber ♦
    Commented Dec 6, 2023 at 20:09

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