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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?

lone_wolf
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  • Related: https://mathoverflow.net/questions/66763/tight-bounds-on-probability-of-sum-of-laplace-random-variables – Galen Dec 06 '23 at 15:44
  • Since they are independent, you can convolve their distributions. Convolution theorem might be handy. – Galen Dec 06 '23 at 15:47
  • 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 https://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? – whuber Dec 06 '23 at 16:06
  • @Galen Then I would have to convolve these N+1 Laplacian distributions right? That will be a complex distribution I think. – lone_wolf Dec 06 '23 at 16:18
  • @wolf_pack_32 That's right; it involves multiple convolutions which gets complicated. I would first explore using a computer algebra system on some chosen values for $N$, and then consider whether those results help me guess what the expansion would look like for the $N+1$ case. – Galen Dec 06 '23 at 16:31
  • For large $N$, combine this approach with this approach. – jblood94 Dec 06 '23 at 17:24
  • I wonder if someone has worked out an error bound between the distribution and large $N$ approximation. – Galen Dec 06 '23 at 18:08
  • @whuber That's helpful. Yes, I agree my approximation is bad. So, do I have to factor the ranges of t as well in the C.F? – lone_wolf Dec 06 '23 at 18:14
  • The range of $t$ plays no role. Because this cf is analytic at the origin, its germ there -- essentially, the entire suite of its derivatives at zero -- determines the distribution. The values of the cf for nonzero $t$ are otherwise irrelevant. What matters are the magnitudes of the coefficients of those derivatives: the powers of $t,$ multiplied by suitable factorials. – whuber Dec 06 '23 at 18:21
  • If I understand you correctly, you ought to say "$N$ independent random variables with a Laplacian distribution with $\mu=0$ and $b=\sigma^2/2$" rather than "$N$ Laplacian distributions with..." etc. – Michael Hardy Dec 06 '23 at 19:05
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    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. – whuber Dec 06 '23 at 20:09

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