To find the covariance of the given joint continuous distribution, we need to calculate it using the formula:

[ \text{Cov}(X, Y) = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y] ]

Here are the steps:

1. Find $\mathbb{E}[X]$:
   The marginal probability density function (PDF) of $X$ is obtained by integrating $f(x, y)$ over $y$:
   $f_X(x) = \int_0^\infty f(x, y) \, dy$
   Then, calculate $\mathbb{E}[X] = \int_0^\infty x f_X(x) \, dx$.

2. Find $\mathbb{E}[Y]$:
   Similarly, the marginal PDF of $Y$ is obtained by integrating $f(x, y)$ over $x$:
   $f_Y(y) = \int_0^\infty f(x, y) \, dx$
   Then, calculate $\mathbb{E}[Y] = \int_0^\infty y f_Y(y) \, dy$.

3. Find $\mathbb{E}[XY]$:
   This is computed directly from the joint PDF:
   $\mathbb{E}[XY] = \int_0^\infty \int_0^\infty xy f(x, y) \, dx \, dy$

4. Combine the results:
   Substitute the values of $\mathbb{E}[XY]$, $\mathbb{E}[X]$, and $\mathbb{E}[Y]$ into the covariance formula.

Let’s compute these step-by-step.

Step 1: Marginal PDF of $X$

Given $f(x, y) = x e^{-(x + y)}$ for $x > 0, y > 0$, $f_X(x) = \int_0^\infty x e^{-(x + y)} \, dy = x e^{-x} \int_0^\infty e^{-y} \, dy$ The integral $\int_0^\infty e^{-y} \, dy = 1$, so: $f_X(x) = x e^{-x}, \quad x > 0$

Step 2: Marginal PDF of $Y$

$f_Y(y) = \int_0^\infty x e^{-(x + y)} \, dx$ Using a similar approach: $f_Y(y) = e^{-y} \int_0^\infty x e^{-x} \, dx$ Here, $\int_0^\infty x e^{-x} \, dx = 1$, so: $f_Y(y) = e^{-y}, \quad y > 0$

Step 3: $\mathbb{E}[X]$ and $\mathbb{E}[Y]$

$\mathbb{E}[X]$ is calculated as: $\mathbb{E}[X] = \int_0^\infty x f_X(x) \, dx = \int_0^\infty x^2 e^{-x} \, dx$ Here, $\int_0^\infty x^2 e^{-x} \, dx = 2$ (gamma function), so: $\mathbb{E}[X] = 2$

$\mathbb{E}[Y]$ is similarly: $\mathbb{E}[Y] = \int_0^\infty y f_Y(y) \, dy = \int_0^\infty y e^{-y} \, dy$ Here, $\int_0^\infty y e^{-y} \, dy = 1$, so: $\mathbb{E}[Y] = 1$

Step 4: $\mathbb{E}[XY]$

$\mathbb{E}[XY] = \int_0^\infty \int_0^\infty xy \cdot x e^{-(x + y)} \, dx \, dy$ Simplify the expression: $\mathbb{E}[XY] = \int_0^\infty \int_0^\infty x^2 y e^{-(x + y)} \, dx \, dy$ Separate the integrals: $\mathbb{E}[XY] = \int_0^\infty x^2 e^{-x} \, dx \cdot \int_0^\infty y e^{-y} \, dy$ We know $\int_0^\infty x^2 e^{-x} \, dx = 2$ and $\int_0^\infty y e^{-y} \, dy = 1$, so: $\mathbb{E}[XY] = 2 \cdot 1 = 2$

Step 5: Covariance

[ \text{Cov}(X, Y) = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y] ] Substitute the values: $\text{Cov}(X, Y) = 2 - (2)(1) = 0$

Final Answer:

The covariance of $X$ and $Y$ is: $\boxed{0}$

Would you like further details or clarification? 😊

Here are 5 related questions:

1. How do you calculate the variance of a single variable from its marginal distribution?
2. Why does the covariance of $X$ and $Y$ turn out to be zero in this case?
3. Can covariance be used to determine independence? Why or why not?
4. How would the calculation change if the joint PDF were different, say $x^2 e^{-(x + y)}$?
5. How is the gamma function used to evaluate integrals like $\int_0^\infty x^n e^{-x} \, dx$?

Tip: Covariance only measures linear relationships; always check if $X$ and $Y$ are truly independent using their joint and marginal distributions.