Question

If X and Y are i.i.d r.v.s with 〖fx (x)=α∙e〗^(-αx) u(x) and 〖fY (y)=α∙e〗^(-αy) u(y). Find the pdf fz (z) for Z=2X+Y.

Answer 1

Answer on Question #48727 – Math – Statistics and Probability

If $X$ and $Y$ are i.i.d r.v.s with $f_{X}(x) = \alpha \cdot e^{-\alpha x}$, $u(x)$ and $f_{Y}(y) = \alpha \cdot e^{-\alpha y}u(y)$. Find the pdf $f_{Z}(z)$ for

$$Z = 2X + Y.$$

Solution

$$F_{2X}(t) = P(2X < t) = P\left(X < \frac{t}{2}\right) = F_{X}\left(\frac{t}{2}\right),$$

Find probability distribution function of $2X$:

$$f_{2X}(t) = \frac{d}{dt} F_{2X}(t) = \frac{d}{dt} F_{X}\left(\frac{t}{2}\right) = \frac{1}{2} f_{X}\left(\frac{t}{2}\right) = \frac{\alpha}{2} \cdot e^{-\frac{\alpha t}{2}} u\left(\frac{t}{2}\right), \text{ where } u(t) = 1 \text{ when } t \geq 0;$$ $$u(t) = 0 \text{ when } t < 0.$$

The probability density function of the sum of two independent random variables $2X$ and $Y$, each of which has a probability density function, is the convolution of their separate density functions:

$$f_{Z}(z) = f_{2X + Y}(z) = \int_{-\infty}^{+\infty} f_{2X}(x) f_{Y}(z - x) dx = \int_{-\infty}^{+\infty} \frac{\alpha}{2} \cdot e^{-\frac{\alpha x}{2}} u\left(\frac{x}{2}\right) \alpha \cdot e^{-\alpha(z - x)} u(z - x)) dx = \alpha \frac{\alpha^{2}}{2} \int_{0}^{z} e^{-\alpha z + \frac{\alpha x}{2}} dx = \alpha e^{-\alpha z} \left. e^{\frac{\alpha x}{2}} \right|_{0}^{z} u(z) = \alpha e^{-\alpha z} \left(e^{\frac{\alpha z}{2}} - 1\right) u(z) = \alpha e^{-\frac{\alpha z}{2}} \left(1 - e^{-\frac{\alpha z}{2}}\right) u(z).$$

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