We are given the joint density function of $X$ and $Y$:

$f(x,y) = tx e^{-y}, \quad 0 < x < 2, \quad y > 0$

and $f(x, y) = 0$ elsewhere.

We need to find the distribution of $Z = X + Y$. This involves finding the probability density function (PDF) of $Z$, denoted as $f_Z(z)$, which can be computed as:

$f_Z(z) = \int_{-\infty}^{\infty} f_X(x) f_Y(z - x) \, dx$

Since $X$ and $Y$ are jointly distributed, we use:

$f_Z(z) = \int_0^2 f(x, z - x) \, dx, \quad \text{for} \quad 0 < x < 2, \quad y > 0.$

Substituting $f(x, y)$:

$f_Z(z) = \int_0^2 x e^{-(z-x)} \, dx, \quad \text{for} \quad z > 0.$

Now, solving the integral:

$f_Z(z) = e^{-z} \int_0^2 x e^x \, dx.$

Using integration by parts, let:

• $u = x$, so $du = dx$.
• $dv = e^x dx$, so $v = e^x$.

Applying integration by parts:

$\int x e^x dx = x e^x - \int e^x dx = x e^x - e^x.$

Evaluating from 0 to 2:

$\left[ x e^x - e^x \right]_0^2 = (2e^2 - e^2) - (0 - e^0) = e^2 - 1.$

Thus,

$f_Z(z) = e^{-z} (e^2 - 1).$

For $z \leq 0$, $f_Z(z) = 0$, so the final PDF is:

\begin{cases} (e^2 - 1)e^{-z}, & z > 0, \\ 0, & \text{otherwise}. \end{cases}$$ This is the exponential distribution with a rate parameter of 1 and a scaling factor $$e^2 - 1$$. Would you like a deeper explanation of any step? ### Related Questions: 1. How do you find the marginal density function of $$X$$ from the given joint density function? 2. What is the expected value $$E[X+Y]$$ using the derived distribution? 3. How do you find the cumulative distribution function (CDF) of $$Z = X + Y$$? 4. Can we verify that $$f_Z(z)$$ is a valid probability density function? 5. What is the variance of $$X+Y$$ given the joint density function? #### Tip: When computing the distribution of a sum of two random variables, consider convolution when they are independent, but joint integration when they are dependent.