We give the solution to the following problem. Suppose that the joint probability density function X and Y is given as f(x,y)=e^-{x+y}. Then find:

- P(X=Y)
- P(min(X,Y)>1/2)
- P(X ≤ Y)
- the marginal density function of Y.
- E[XY]

**Solution:**

We are given that X and Y have the following joint PDF, f(x,y) = \begin{cases} e^{-(x+y)} & \text{ if } x > 0, y > 0 \\ 0 & \text{ otherwise} \end{cases}

- We define a new variable Z = X – Y. Since X and Y are random variables so is Z. Now, P(X=Y) = P(X-Y=0) = P(Z=0). But we know that the probability of a random variable taking a simple point value is always zero. Therefore, (Z=0) = 0 which is the required answer.
- The minimum value of X and Y is greater than 1/2 if and only if both X and Y themselves take values greater than 1/2. This is because if either X and Y were less than 1/2 then their minimum value would definitely be less than 1/2. Therefore we have that, P(min(X,Y)>1/2) = P(X>1/2,Y>1/2) = \int_{\frac{1}{2}}^{\infty}\int_{\frac{1}{2}}^{\infty}f(x,y) dx dy \\ = \int_{1/2}^{\infty}\int_{1/2}^{\infty}e^{-(x+y)} dx dy \\ = \int_{1/2}^{\infty}\int_{1/2}^{\infty}e^{-x}e^{-y} dx dy \\ = (\int_{1/2}^{\infty}e^{-x}dx)\times (\int_{1/2}^{\infty}e^{-y} dy) \\ = [\frac{e^{-x}}{-1}]_{1/2}^{\infty}\times [\frac{e^{-y}}{-1}]_{1/2}^{\infty} \\ = e^{-1/2}\times e^{-1/2} = e^{-1} = 0.3679.
- In order to calculate P(X ≤ Y), we let X vary between 0 and Y, whereas Y varies freely between 0 to infinity. Therefore we have that, P(X\leq Y) = \int_{0}^{\infty}\int_{0}^{y}f(x,y) dx dy \\ = \int_{0}^{\infty}\int_{0}^{y} e^{-(x+y)} dx dy \\ = \int_{0}^{\infty}(\int_{0}^{y}e^{-x}dx )e^{-y} dy \\ = \int_{0}^{\infty}( 1- e^{-y} )e^{-y} dy \\ = \int_{0}^{\infty}(e^{-y} - e^{-2y}) dy \\ = 1 - 1/2 = 0.5
- In order to find the marginal probability density function of Y, we integrate the joint PDF over the random variable X. f(x) = \int_{0}^{\infty}f(x,y) dx \\ = \int_{0}^{\infty} e^{-(x+y)} dx \\ = e^{-y} \int_{0}^{\infty}e^{-x}dx \\ = e^{-y}\times 1 \\ = e^{-y} So the marginal density function of Y is, f(y) = \begin{cases} e^{-y} & \text{ if } y > 0 \\ 0 & \text{ otherwise} \end{cases}
- We use the formula, E[XY] = \int_{0}^{\infty}\int_{0}^{\infty}xyf(x,y) dx dy \\ = (\int_{0}^{\infty}xe^{-x}dx)\times (\int_{0}^{\infty}ye^{-y} dy) \\ = [-xe^{-x} + e^{-x}]_0^{\infty}\times [-ye^{-y} + e^{-y}]_0^{\infty} = 1