Sum Pdf 7Facd1

1. **State the problem:** We have two independent and identically distributed random variables $X$ and $Y$ with pdf $f(x) = e^{2-x}$ for $x > 2$ and 0 otherwise. We want to find the pdf of $Z = X + Y$. 2. **Recall the formula for the sum of independent random variables:** $$f_Z(z) = \int_{-\infty}^\infty f_X(x) f_Y(z - x) \, dx$$ Since $X$ and $Y$ are independent and identically distributed, $f_X = f_Y = f$. 3. **Determine the support of $Z$:** Since $X > 2$ and $Y > 2$, the minimum value of $Z = X + Y$ is $4$. So $f_Z(z) = 0$ for $z \leq 4$. 4. **Set up the convolution integral for $z > 4$:** $$f_Z(z) = \int_{2}^{z-2} f(x) f(z - x) \, dx$$ Because $f(x) = e^{2-x}$ for $x > 2$ and 0 otherwise, the limits are from $2$ to $z-2$. 5. **Substitute $f(x)$ and $f(z-x)$:** $$f_Z(z) = \int_{2}^{z-2} e^{2-x} e^{2-(z-x)} \, dx = \int_{2}^{z-2} e^{2-x + 2 - z + x} \, dx = \int_{2}^{z-2} e^{4 - z} \, dx$$ 6. **Simplify the integrand:** $$e^{4 - z}$$ is constant with respect to $x$, so $$f_Z(z) = e^{4 - z} \int_{2}^{z-2} dx = e^{4 - z} (z - 2 - 2) = e^{4 - z} (z - 4)$$ 7. **Write the final pdf of $Z$:** $$f_Z(z) = \begin{cases} (z - 4) e^{4 - z} & z > 4, \\ 0 & \text{otherwise} \end{cases}$$ **Answer:** The pdf of $Z = X + Y$ is $$f_Z(z) = (z - 4) e^{4 - z} \text{ for } z > 4, \text{ and } 0 \text{ otherwise.}$$