# Calculating pdf for sum of two random variables

I have a random process Z(t).This function is defined as equal to A(t)+B. Well, It's clear that to obtain the pdf of Z(t), I need to take the convolution but I am not getting it how in the solution below the person has calculated pdf for A'(t).Why there are two |t| in denominator.

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If $X$ is a continuous random variable with probability density function $f_X(x)$, then, for $c\neq 0$, so is $cX$ a continuous random variable with probability density function $$f_{cX}(a)=\frac{1}{|c|}f_X\left(\frac{a}{c}\right)$$ – Dilip Sarwate Jan 13 '13 at 14:25

Your question "Why are there two $|t|$ in the denominator?" is likely due to careless reading of the material you posted. There are two $t$'s in the denominators but only one of them is the absolute value $|t|$.
Suppose that $X$ is a continuous random variable and that $Y = cX$ for some nonzero real number $c$. Then, for $c > 0$, $$F_Y(y) = P\{Y \leq y\} = P\{cX \leq y\} = P\left\{X \leq \frac{y}{c}\right\} = F_X\left(\frac{y}{c}\right).$$ The chain rule for differentiation thus gives $$f_Y(y) = \frac{\mathrm d}{\mathrm dy}F_Y(y) = \frac{\mathrm d}{\mathrm dy}F_X\left(\frac{y}{c}\right) = f_X\left(\frac{y}{c}\right)\times\frac{1}{c} = \frac{1}{|c|}f_X\left(\frac{y}{c}\right).$$ For $c < 0$, a similar calculation yields $$F_Y(y) = P\left\{X \geq \frac{y}{c}\right\} = 1 - F_X\left(\frac{y}{c}\right)$$ since $X$ is a continuous random variable which leads to $$f_Y(y) = -f_X\left(\frac{y}{c}\right)\times\frac{1}{c} = \frac{1}{|c|}f_X\left(\frac{y}{c}\right).$$
Apart from the mathematical derivation above, there is an intuitive method of arriving at the same answer. Suppose that $c > 0$. The value of $Y = cX$ is $c$ times "larger" than the value of $X$, that is, we have stretched (compressed if $0 < c < 1$) the horizontal axis by a factor of $c$. It would seem reasonable to have the value of $f_X(3)$, say, be the same as the value of $f_Y(3c)$, or, put the other way around, $f_Y(y) = f_X\left(\frac{y}{c}\right)$, sounds about right, no? Well, not quite because the function we end up with must be a density with "total area under the curve" equal to $1$. So if we have stretched the horizontal axis by a factor of $c$, the height of the function must be reduced by the same factor to keep the area the same. In short, for $c > 0$, the answer must be $$f_Y(y) = \frac{1}{c}f_X\left(\frac{y}{c}\right).$$ I will leave it for you to work out that for $c < 0$, $f_Y$ is obtained from $f_X$ by stretching the horizontal axis, compressing the vertical axis, and "flipping over" the function with respect to the vertical axis.