Sum of independent exponential random variables with common parameter

Question

Let $X$ and $Y$ be independent exponential random random variables with common parameter $\lambda$ and let $Z = X + Y$. Find $f_Z(z)$.

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My approach:

Step 1: $$F_Z(z) = P(X + Y \leq Z) = \int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{z-x}f_{X,Y}(x,y)dydx$$

Step 2: $$f_Z(z) = \frac{d}{dz}F_Z(z) = \int\limits_{-\infty}^{\infty}\frac{d}{dz}[\int\limits_{-\infty}^{z-x}f_{X,Y}(x,y)dy]dx = \int\limits_{-\infty}^{\infty}f_{X,Y}(x,z-x)dx$$

Step 3: Since the variables are independent: $$f_Z(z) = \int\limits_{-\infty}^{\infty}f_{X}(x)*f_{Y}(z-x)dx$$

Step 4: Using exponential function formula: $\lambda e^{-\lambda x}$ and that lower bound for exponential is 0 to infinity: $$f_Z(z) = \int\limits_{0}^{\infty} \lambda e^{-\lambda x}* \lambda e^{-\lambda (z-x)} dx = \lambda ^2\int\limits_{0}^{\infty} e^{-\lambda z}dx$$

I think I took a long turn somewhere because I'm getting an integral of a constant as my result. Where did I go wrong?

Answer 1

The mistake is in the upper bound in the very last integral. It should be $z$ because $x+y\leq z$ and since $Y$ is exponentially distributed it is nonnegative. Then we have

$$f_Z(z)=\lambda^2\int_0^ze^{-\lambda z}dx=\lambda^2 ze^{-\lambda z}$$

Note that this is the answer we expect since the sum of two exponential with parameter $\lambda$ is distributed as $\Gamma(2,\lambda)$ which has density

$$f_Z(z)=\frac{\lambda^2z^{2-1}e^{-\lambda z}}{\Gamma(2)}$$

Answer 2

For $x\gt z$ the integrand $=0$ since $f_Y(y)=0$ for $y\lt 0$.

$f_Z(z)=\lambda^2 e^{-\lambda z}\int\limits_0^zdx=z\lambda^2 e^{-\lambda z}$.