Probability Theory – Problem 52: Find the distribution of ;

The random variables $X$ and $Y$ have joint density function $p(x,\ y)=x e^{-x(y+1)}$. (1) Find the distribution of $X Y$; (2) Find the distribution of $X+X Y$.

We first determine the support of the random vector $(X, Y)$, i.e., the region where the joint density $p(x, y)$ is positive. Since $p(x, y) = x e^{-x(y+1)}$, positivity requires $x>0$. To ensure the density integrates to 1 over the entire plane, we determine the range of $y$. The support is $x>0, y>0$. Verification: $$\int_0^\infty \int_0^\infty x e^{-x(y+1)} \,\mathrm{d}y \,\mathrm{d}x = \int_0^\infty x e^{-x} \left( \int_0^\infty e^{-xy} \,\mathrm{d}y \right) \,\mathrm{d}x$$ Computing the inner integral: $$\int_0^\infty e^{-xy} \,\mathrm{d}y = \left[ -\frac{1}{x} e^{-xy} \right]_{y=0}^{y=\infty} = 0 - \left(-\frac{1}{x}\right) = \frac{1}{x}$$ Substituting into the outer integral: $$\int_0^\infty x e^{-x} \left( \frac{1}{x} \right) \,\mathrm{d}x = \int_0^\infty e^{-x} \,\mathrm{d}x = \left[ -e^{-x} \right]_0^\infty = 0 - (-1) = 1$$ The integral equals 1, so the support of $(X, Y)$ is $\{ (x,y) | x>0, y>0 \}$. (1) Find the distribution of $Z = XY$. Using the CDF method. Since $X>0, Y>0$, we have $Z>0$. For $z \le 0$, $F_Z(z) = P(Z \le z) = 0$. For $z > 0$, the CDF is: $$F_Z(z) = P(Z \le z) = P(XY \le z) = \iint_{xy \le z, x>0, y>0} x e^{-x(y+1)} \,\mathrm{d}x \,\mathrm{d}y$$ Expressing the integration region as $0 < y \le z/x, 0 < x < \infty$: $$F_Z(z) = \int_0^\infty \left( \int_0^{z/x} x e^{-x(y+1)} \,\mathrm{d}y \right) \,\mathrm{d}x$$ Computing the inner integral with respect to $y$: $$\int_0^{z/x} x e^{-x(y+1)} \,\mathrm{d}y = x e^{-x} \int_0^{z/x} e^{-xy} \,\mathrm{d}y = x e^{-x} \left[ -\frac{1}{x} e^{-xy} \right]_0^{z/x}$$ $$= x e^{-x} \left( -\frac{1}{x} e^{-z} + \frac{1}{x} \right) = e^{-x}(1 - e^{-z})$$ Substituting into the outer integral: $$F_Z(z) = \int_0^\infty e^{-x}(1 - e^{-z}) \,\mathrm{d}x = (1 - e^{-z}) \int_0^\infty e^{-x} \,\mathrm{d}x = (1 - e^{-z}) \cdot 1 = 1 - e^{-z}$$ This is the CDF of the exponential distribution with parameter 1. Differentiating $F_Z(z)$ yields the PDF $p_Z(z)$ of $Z=XY$: $$p_Z(z) = \frac{\mathrm{d}}{\mathrm{d}z} (1 - e^{-z}) = e^{-z}, \quad z>0$$

(2) Find the distribution of $W = X+XY = X(1+Y)$. Again using the CDF method. Since $X>0, Y>0$, we have $W>0$. For $w \le 0$, $F_W(w) = 0$. For $w > 0$, the CDF is: $$F_W(w) = P(W \le w) = P(X(1+Y) \le w) = \iint_{x(1+y) \le w, x>0, y>0} x e^{-x(y+1)} \,\mathrm{d}x \,\mathrm{d}y$$ The integration region can be expressed as $0 < y \le \frac{w}{x} - 1$. For $y>0$, we need $\frac{w}{x} - 1 > 0$, i.e., $x < w$. So the integration limits are $0 < x < w$: $$F_W(w) = \int_0^w \left( \int_0^{\frac{w}{x}-1} x e^{-x(y+1)} \,\mathrm{d}y \right) \,\mathrm{d}x$$ Computing the inner integral with respect to $y$: $$\int_0^{\frac{w}{x}-1} x e^{-x(y+1)} \,\mathrm{d}y = x e^{-x} \int_0^{\frac{w}{x}-1} e^{-xy} \,\mathrm{d}y = x e^{-x} \left[ -\frac{1}{x} e^{-xy} \right]_0^{\frac{w}{x}-1}$$ $$= x e^{-x} \left( -\frac{1}{x} e^{-x(\frac{w}{x}-1)} + \frac{1}{x} \right) = e^{-x} (1 - e^{-w+x}) = e^{-x} - e^{-w}$$ Substituting into the outer integral: $$F_W(w) = \int_0^w (e^{-x} - e^{-w}) \,\mathrm{d}x = \left[ -e^{-x} \right]_0^w - e^{-w} \left[ x \right]_0^w$$ $$= (-e^{-w} - (-e^0)) - e^{-w}(w-0) = 1 - e^{-w} - w e^{-w} = 1 - (1+w)e^{-w}$$ Differentiating $F_W(w)$ yields the PDF $p_W(w)$ of $W=X+XY$: $$p_W(w) = \frac{\mathrm{d}}{\mathrm{d}w} (1 - (1+w)e^{-w}) = -[1 \cdot e^{-w} + (1+w)(-e^{-w})]$$ $$= -[e^{-w} - e^{-w} - w e^{-w}] = w e^{-w}, \quad w>0$$ This is the density function of the Gamma distribution with parameters $\alpha=2, \beta=1$.

(1) $XY$ follows the exponential distribution with parameter 1, with PDF $p(z) = e^{-z}$ for $z>0$. (2) $X+XY$ follows the Gamma distribution with parameters $\alpha=2, \beta=1$ ($\Gamma(2,1)$), with PDF $p(w) = w e^{-w}$ for $w>0$.