Finding the probability density function of $X=2T_1+T_2$.

Question

I am finding the probability density function $$ X=2T_1+T_2.$$ $T_1,T_2$ are independent with density function of $$f(t)=e^{-t}, t>0.$$ So what I did is find the $f(x)$ by using this integration: $$\int_{0}^{\infty} e^{-t_1}\cdot e^{-(x-2t_1)} dt_1.$$ May I know why did I go wrong, thx so much.

Answer 1

There is absolutely nothing wrong with your working except the upper bound of your integration,

$X = 2T_1 + T_2 \implies 0 \leq t_1 \leq \frac{x}{2}$

So the integral becomes,

$\displaystyle \int_{0}^{x/2} e^{-t} e^{-(x-2t)} \ dt = \int_{0}^{x/2} e^{-x+t} \ dt = e^{-x/2} - e^{-x}$

So, $\displaystyle f_X(x) = e^{-x/2} - e^{-x}, x \geq 0$

Answer 2

For $X=2T_1+T_2$, you should evaluate $$F(x)=\iint_{2t_1+t_2 \leq x, t_1> 0,t_2>0}e^{-t_1}\cdot e^{-t_2}dt_1dt_2=1-2e^{-x/2}+e^{-x}.$$ Then the density function is $f(x)=F'(x)=e^{-x/2}-e^{-x}$.

Answer 3

The density of $2T_1$ is $\frac 1 2 e^{-x/2}, x>0$ so the density of $X$ is $\int_0^{x} \frac 12 e^{-(x-y)/2} e^{-y}dy$ for $x>0$.