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Given independent and identically distributed random variables $X_1, X_2, \dots, X_n$, each of them has the same p.d.f $f(x) = Pr(X = x)$ on support $(a, b)$.

How do I find the pdf or cdf of $Y = \sum_{i = 1}^n a_iX_i$, where $1 \le i \le n$ and $a_i$'s are constants?

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The pdf of $\alpha_iX_i$ is $\frac{1}{|\alpha_i|}f(\frac{x}{\alpha_i})$ which we shall denote $g_i(x)$. Then since the $\alpha_iX_i$ are independent, the pdf of the sum Y is the convolution of the $g_i$:

$$f_Y(y) = (g_1 * g_2 * g_3 * \dots *g_n)(y)$$

where * means convolution.

$$(g_i * g_j)(y) = \int_{-\infty}^{\infty}g_i(t)g_j(y-t)dt$$

for continuous random variables, while for discrete random variables it would be

$$(g_i * g_j)[n] = \sum_{k = -\infty}^{\infty}g_i[k]g_j[n-k].$$

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  • $\begingroup$ If the OP really means to say "p.d.f $f(x) = \Pr(X=x)$", then $X$ should be construed as a discrete random variable and not a continuous random variable with probability density function $f(x)$. $\endgroup$ – Dilip Sarwate Jun 15 '15 at 2:56
  • $\begingroup$ He said it had support on (a,b) which implies it's continuous, but the convolution integral would still be correct for the discrete case where the pdf is given in terms of delta functions. For the continuous case, Pr(X=x) would refer to a differential. I added the summation definition of convolution though. I saw there was a comment that said the original notation was clear, but it's gone now. $\endgroup$ – BruceZ Jun 15 '15 at 3:19

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