What is the probability density function of $X_1+X_2$?

Question

Given $X_1 \sim \exp(\lambda_1)$ and $X_2 \sim \exp(\lambda_2)$, and that they are independent, how can I calculate the probability density function of $X_1+X_2$?

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I tried to define $Z=X_1+X_2$ and then: $f_Z(z)=\int_{-\infty}^\infty f_{Z,X_1}(z,x) \, dx = \int_0^\infty f_{Z,X_1}(z,x) \, dx$.
And I don't know how to continue from this point.

Answer 1

$$f_Z(z)=\int f_{X_1}(x)f_{X_2}(z-x)dx$$

Note that in your case the RHS has integrand $0$ if $z\leq0$ so that $f_Z(z)=0$ if $z\leq0$.

For $z>0$ we have:$$f_Z(z)=\int f_{X_1}(x)f_{X_2}(z-x)dx=\int_0^{z}f_{X_1}(x)f_{X_2}(z-x)dx$$

Work this out yourself.

Answer 2

HINT:

Assuming independence.

We are given that the P.D.F. of $X$ is $f_X(x)=\lambda e^{-\lambda x},\, x\ge0$ and the P.D.F. of $Y$ is $f_Y(Y)=\lambda e^{-\lambda y},\, y\ge0$.

Then using convolution, $$\begin{align}f_{X+Y}(x+y)&=\int_{-\infty}^\infty f_X(x+y-y)f_Y(y)\,dy\\&=\int_0^{x+y}\lambda e^{-\lambda(x+y)}\lambda e^{-\lambda y}\,dy\end{align}$$ which should now to easy to integrate.

Answer 3

Just as has been pointed out by the other answers, you can simply calculate the pdf for $X_1 + X_2$ by using the principle of convolution. In fact, in general one can show that if $X_1,X_2,...X_n$ are i.i.d variables with exponential distribution with parameter $\lambda$ then $S = \sum_{k=1}^{n}X_k \sim \Gamma (n,\lambda)$.