Pdf of $T_1+T_2$ and $T_2-T_1$, separately, where $T_i\sim Exp(\lambda_i)$ independently.

Question

Let $T_i$ be independent, exponentially distributed rvs, such that $T_i\sim Exp(\lambda_i)$. Evaluate the pdf for each one of $T_1+T_2$ and $T_2-T_1$.

Attempt. One thought is to work jointly: set $U=T_2+T_1,~V=T_2-T_1$ and find the joint pdf of $(U,V)$ through the Jacobian determinant. Then I could work separately on $U$ and $V$, using $f_U(u)=\int f_{U,V}(u,v)dv$ and $f_V(v)=\int f_{U,V}(u,v)du$. Is there a quicker way to work? By means of memoryloss properties of total probability theorem maybe?

Thanks in advance!

Answer 1

Let $X=T_1+T_2$ and $Y=T_2-T_1$. If $\lambda_1=\lambda_2$ then $X$ has $\mathsf{Erlang}(2,\lambda_1)$ distribution, with density $$ f_X(t) = \lambda_1(\lambda_1 t)e^{-\lambda_1 t}\cdot\mathsf 1_{(0,\infty)}(t). $$ If $\lambda_1\ne\lambda_2$ then we can compute the density by convolution: \begin{align} f_X(t) &= (f_{T_1}\star f_{T_2})(t)\\ &= \int_0^\infty f_{T_1}(s)f_{T_2}(t-s)\ \mathsf ds\\ &= \int_0^\infty \lambda_1 e^{-\lambda_1 s}\lambda_2 e^{-\lambda_2(t-s))}\ \mathsf ds\\ &= \frac{\lambda_1\lambda_2}{\lambda_1-\lambda_2}\left(e^{-\lambda_2 t}-e^{-\lambda_1 t}\right)\cdot\mathsf 1_{(0,\infty)}(t). \end{align}

For the distribution of $Y$ I will defer to this answer: pdf of the difference of two exponentially distributed random variables

Answer 2

Sure, if you have a reason to find the joint pdf for $(U,V)$, you can do that.

Otherwise, it is possibly easier to find the joint pdf for ($T_1,U$) and ($T_1,V)$, then find the marginals from those.

$$f_U(u)=\int_\Bbb R f_{T_1,U}(t,u)~\mathsf d t = \int_\Bbb R f_{T_1,T_2}(t,u-t)~\mathsf d t $$

$$f_V(v)=\int_\Bbb R f_{T_1,V}(t,v)~\mathsf d t = \int_\Bbb R f_{T_1,T_2}(t,v+t)~\mathsf d t $$