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This question is about finding the probability density function of square of the sum of $n$ independent Rayleigh distributed random variables $$S= \left(\sum_{i=1}^n X_i\right)^2$$

$$X_i \sim \frac{x}{M_i^2} e^{-\frac{x^2}{2M_i^2}}$$

The paper I am reading says that $$\mathbb{P}\bigl(S\leq \theta\bigl)= \int\limits_{\sum_{i=1}^nx_i \leq \theta} \prod_{i=1}^n \frac{x_i}{M_i^2} e^{-\frac{x_i^2}{2M_i^2}} dx$$

I know that the sum of independent random variables should have a pdf that is obtained after convolution, how come the distribution is the product of the individual distributions?

Many thanks

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The formula should be $$\mathbb{P}\bigl(S\leqslant \theta\bigl)= \int\limits_{\left(\sum_{i=1}^nx_i\right)^2 \leqslant \theta} \prod_{i=1}^n \frac{x_i}{M_i^2} e^{-\frac{x_i^2}{2M_i^2}} \mathrm dx.$$

This can be obtained in the following way: if $(X_i)_{1\leqslant i\leqslant n}$ is a collection of $n$ independent random variable and $X_i$ has density $f_i$, then a density of $(X_1,\dots,X_n)$ is given by $$f_{(X_1,\dots,X_n)}(x_1,\dots,x_n)=\prod\limits_{i=1}^nf_i(x_i).$$

If $(X_1,\dots,X_n)$ is a random vector with density $f$, then for each bounded measurable function $\varphi\colon\mathbb R^n\to\mathbb R$, $$\mathbb E[\varphi(X_1,\dots,X_n)]=\int_{\mathbb R^n}\varphi\left(x_1,\dots,x_n\right)f(x_1,\dots,x_n)\mathrm dx.$$ Use this when $\varphi$ is the characteristic function of the set $\left\{(x_1,\dots,x_n)\in\mathbb R^n,\left(\sum_{i=1}^nx_i\right)^2 \leqslant \theta\right\}$.

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  • $\begingroup$ And is this case we have something of the form $$\mathbb{E}[1_{S\leq\theta}]$$? $\endgroup$
    – Tyrone
    Nov 13, 2014 at 15:18
  • $\begingroup$ Isn't is the same as the quantity $\mathbb P\{S\leqslant \theta\}$? $\endgroup$ Nov 14, 2014 at 9:57
  • $\begingroup$ yes i think so! $\endgroup$
    – Tyrone
    Nov 14, 2014 at 15:52

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