The random variables and have the joint density function . (1) Find the distribution of ; (2) Find the distribution of .

(1) follows an exponential distribution with parameter 1, with probability density function for . (2) follows a Gamma distribution with parameters (i.e., ), with probability density function for .

Probability Theory Solution – Problem 9: Find the distribution of ;

Question

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

Step-by-step solution

We first determine the support of the random vector $(X, Y)$ , i.e., the region where the joint density function $p(x, y)$ is positive. Since $p(x, y) = x e^{-x(y+1)}$ , positivity requires $x > 0$ . To ensure that the density integrates to 1 over the entire plane, we must determine the range of $y$ . The support is $x > 0, y > 0$ . We verify this as follows: $\int_0^\infty \int_0^\infty x e^{-x(y+1)} \,\mathrm{d}y \,\mathrm{d}x = \int_0^\infty x e^{-x} \left( \int_0^\infty e^{-xy} \,\mathrm{d}y \right) \,\mathrm{d}x$ Evaluating 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, confirming that the support of $(X, Y)$ is $\{ (x,y) | x>0, y>0 \}$ . Step 1. Compute the distribution of $Z = XY$ . We use the CDF method. Since $X > 0$ and $Y > 0$ , 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 region of integration as $0 < y \le z/x,\, 0 < x < \infty$ : $F_Z(z) = \int_0^\infty \left( \int_0^{z/x} x e^{-x(y+1)} \,\mathrm{d}y \right) \,\mathrm{d}x$ Evaluating 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 an exponential distribution with parameter 1. Differentiating $F_Z(z)$ yields the PDF of $Z = XY$ : $p_Z(z) = \frac{\mathrm{d}}{\mathrm{d}z} (1 - e^{-z}) = e^{-z}, \quad z>0$

Step 2. Compute the distribution of $W = X + XY = X(1+Y)$ . We again use the CDF method. Since $X > 0$ and $Y > 0$ , 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 region of integration can be expressed as $0 < y \le \frac{w}{x} - 1$ . For $y > 0$ , we need $\frac{w}{x} - 1 > 0$ , i.e., $x < w$ . Hence the limits of integration are $0 < x < w$ : $F_W(w) = \int_0^w \left( \int_0^{\frac{w}{x}-1} x e^{-x(y+1)} \,\mathrm{d}y \right) \,\mathrm{d}x$ Evaluating 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 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 of a Gamma distribution with parameters $\alpha=2$ and $\beta=1$ .

Final answer

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

Marking scheme

This is an undergraduate mathematics marking scheme (7 points total). The scheme is based on the official solution logic and is also compatible with equivalent approaches such as the change-of-variables method or structural/distributional-property arguments.

1. Checkpoints (max 7 pts total)

Part I: Finding the distribution of $XY$ (Max 3 pts)

Score exactly one chain:

*Chain A: CDF method / Change-of-variables method*

Setting up the integral or transformation [1 pt]: Correctly writing the double integral defining the CDF (with correct limits of integration, e.g., $0 < y \le z/x$ ) or correctly writing the joint density under a change of variables together with the Jacobian determinant $|J|=1/x$ .
Carrying out the computation [1 pt]: Correctly evaluating the inner integral or the marginalization integral to obtain an intermediate result (e.g., $F_Z(z) = 1 - e^{-z}$ ).
Final result [1 pt]: Differentiating the CDF or simplifying to obtain the correct density $p(z) = e^{-z}$ (with the domain $z>0$ specified).

*Chain B: Distributional-property method (structural recognition)*

Identifying the distributions [1 pt]: Correctly identifying the marginal distribution $X \sim Exp(1)$ and the conditional distribution $Y|X \sim Exp(x)$ .
Scaling argument [1 pt]: Using the scaling property of the exponential distribution ( $T \sim Exp(\lambda) \implies cT \sim Exp(\lambda/c)$ ) to argue that $Z = X \cdot (Y|X)$ follows $Exp(1)$ .
Final result [1 pt]: Explicitly stating the density $p(z) = e^{-z}$ (with the domain $z>0$ specified).

Part II: Finding the distribution of $X+XY$ (Max 4 pts)

Score exactly one chain:

*Chain A: CDF method / Change-of-variables method*

Setting up the integral or transformation [1 pt]: Correctly formulating the integral expression for $P(X(1+Y) \le w)$ or setting up the change-of-variables framework with $W=X(1+Y)$ .
Determining the critical region [1 pt]: (Key difficulty) Correctly deriving that the upper limit for the variable $x$ is $w$ (i.e., $0 < x < w$ ), because $y = w/x - 1 > 0$ .
*Note: If the upper limit of integration is incorrectly written as $\infty$ instead of $w$ , this checkpoint scores 0.*
Carrying out the computation [1 pt]: Correctly completing the integration to obtain $F_W(w) = 1 - (1+w)e^{-w}$ , or the corresponding marginalization of the joint density.
Final result [1 pt]: Differentiating to obtain the correct density $p(w) = w e^{-w}$ (with the domain $w>0$ specified).

*Chain B: Convolution method (exploiting independence)*

Independence statement [1 pt]: Based on the result of Part I, explicitly stating that $X$ and $XY$ are independent (since the distribution of $XY$ does not depend on $X$ ).
Model identification [1 pt]: Reformulating the problem as the sum of two independent $Exp(1)$ random variables, and invoking the convolution formula or properties of the Gamma distribution.
Computation [1 pt]: Correctly carrying out the convolution integral or writing out the explicit parametric form of the Gamma(2,1) distribution.
Final result [1 pt]: Explicitly stating the density $p(w) = w e^{-w}$ (with the domain $w>0$ specified).

Total (max 7)

2. Zero-credit items

Merely copying the joint density function from the problem statement without setting up any specific integral.
Only verifying that $\iint p(x,y)\, dx\, dy = 1$ without proceeding to compute the required distributions (unless this verification is a necessary step for determining the limits of integration).
Only writing the general definition of a CDF $F(z) = \int_{-\infty}^z p(t)\, dt$ without substituting the specific functions from this problem.

3. Deductions

Missing domain specification (Max -1): The final result does not indicate the range of the variable (e.g., $z>0$ or $w>0$ ), nor introduces an indicator function. At most 1 point deducted across the entire paper.
Logical gap (Max -2): In Part II, failing to discuss the critical constraint $x < w$ yet arriving at the correct final result (e.g., by ad hoc manipulation of the integral to match the answer).
Algebraic error (Max -1): A sign error or minor arithmetic mistake in the differentiation or simple computation steps that does not affect the overall logic.

Probability Theory – Problem 9: Find the distribution of ;