I have 2 exponential random-variables. $T_1\sim\exp(\lambda_1)$ and $T_2\sim \exp(\lambda_2)$.
I want to find $E[T_1|T_1<T_2]$ with integration, however I am not able to set up the proper integral.
First I have to find this probability:
$$P(T_1<T_2)=\int_{0}^\infty\left[\int_{0}^{t_2}\lambda_1e^{-\lambda_1t_1} dt_1\right]\lambda_2e^{-\lambda_2t_2} dt_2=\frac{\lambda_1}{\lambda_1+\lambda_2}, \tag{1}$$
But then I need to find:
$P(T_1 \le t_1 |T_1<T_2)$ (2), which I must differentiate to get:
$f(T_1=t_1|T_1<T_2)$ (3), which I then can integrate to get the desired expectation, like this:
$$E[T_1|T_1<T_2]=\int_{0}^\infty t_1*f(T_1=t_1|T_1<T_2) dt_1 \tag{4}$$
However, finding (2) and (3) is difficult:
I know that: $P(T_1 \le t_1 |T_1<T_2)=\frac{P(T_1<t_1 \cap (T_1<T_2))}{P(T_1<T_2)}(5)$, and the denominator I have from (1), but how do I find the numerator?
Neither (2) nor (3) are necessary. Note that, by definition, $E(T_1;T_1\lt T_2)$ is $$ \int_0^\infty uf_{T_1}(u)\int_u^\infty f_{T_2}(v)\mathrm dv\mathrm du=\int_0^\infty \lambda_1u\mathrm e^{-\lambda_1 u}P(T_2\gt u)\mathrm du. $$ Can you finish this?