Expected payoff based on two independent exponential distributions I am trying to check if I am formulating a problem correctly.
Suppose there are two independent exponential random variables $t_1$ with rate $x$ and $t_2$ with rate $y$. This means that the CDF of $t_1$ is $F(t_1)= 1 - e^{-xt_1}$ for $t_1\geq 0$ and the CDF of $t_2$ is $G(t_2)= 1 - e^{-yt_2}$ for $t_2\geq 0$.
If $t_1 \leq t_2$, the agent gets a flow payoff of 1 from $t_1$ until $t_2$ at which point the flow payoff changes to  $\alpha$ and lasts from $t_2$ until infinity. If instead $t_1 < t_2$, the agent receives a flow payoff of $1-\alpha$ from $t_1$ until infinity and no payoff before $t_1$. Payoffs are discounted exponentially with rate $r$; that means that if, for example, the agent was geting a deterministic flow payoff of $v$ from $\tau$ to $\tau^\prime$, the overall payoff would be $\int_{\tau}^{\tau\prime}ve^{-rs}ds$.
I want to write a formula for the expected overall payoff. I try to break it into two cases for whether $t_1$ is before or after $t_2$. I am not sure if I am correctly formulating the expected value.
The unconditional density of $t_1$ is $f(t_1) = xe^{-xt_1}$. The unconditional density of $t_2$ is $g(t_2) = ye^{-yt_2}$.
$\int_{t_1}^{\infty} ye^{-yt_2} dt_2 = e^{-yt_1}$, so the density of $t_2$ conditional on $t_1 < t_2$ is $$g_{\vert t_1<t_2}(t_2) = \frac{1}{e^{-yt_1}}ye^{yt_2}=ye^{-y(t_2-t_1)}$$
For a fixed $t_1$, the probability that $t_2\leq t_1$ is $G(t_1)$.
Note that with that in place, here is my attempted formula for the expected payoff.
\begin{equation}
\int_0^\infty f(t_1) (1 - G(t_1)) 
\int_{t_1}^{\infty} g_{\vert t_1<t_2}\left(t_2 \right) \left[ \int_{t_1}^{t_2}e^{-rs}(1)ds + \int_{t_2}^{\infty} e^{-rs}(\alpha)ds \right]dt_2 dt_1 \\
+ \int_0^\infty f(t_1) G(t_1) \int_{t_1}^{\infty}e^{-rs}(1-\alpha)dsdt_1
\end{equation}
\begin{equation}
\int_0^\infty \left( xe^{-xt_1} \right) \left(1 - \left(1 - e^{-yt_1} \right) \right) 
\int_{t_1}^{\infty} ye^{-y(t_2 - t_1)} \left[ \int_{t_1}^{t_2}e^{-rs}(1)ds + \int_{t_2}^{\infty} e^{-rs}(\alpha)ds \right] dt_2 dt_1 \\
+ \int_0^\infty xe^{-xt_1} \left(1-e^{-yt_1} \right) \int_{t_1}^{\infty}e^{-rs}(1-\alpha)dsdt_1
\end{equation}
Is this correct?
Thanks
 A: First, the payoff can be computed in general as
$$\int^{\tau'}_{\tau} ve^{-rs}ds = \frac{v}{r}\int^{\tau'}_{\tau} re^{-rs}ds = \frac{v}{r}\left(e^{-r\tau}-e^{-r\tau'}\right).$$
Thus, the agent's total (random) payoff becomes
\begin{eqnarray*}
P &=& 1_{\left\{T_1 < T_2\right\}}\left[\frac{1}{r}\left(e^{-rT_1}-e^{-rT_2}\right)+\frac{\alpha}{r}e^{-rT_2}\right] + 1_{\left\{T_1 \geq T_2\right\}}\frac{1-\alpha}{r}e^{-rT_1}\\
&=& 1_{\left\{T_1 < T_2\right\}}\frac{1}{r} e^{-rT_1} - 1_{\left\{T_1 < T_2\right\}}\frac{1-\alpha}{r}e^{-r T_2} + 1_{\left\{T_1 \geq T_2\right\}}\frac{1-\alpha}{r}e^{-rT_1}
\end{eqnarray*}
The expectation of each term can be represented as a double integral over some part of $\left[0,\infty\right)^2$. It looks like a tedious computation, but should be doable (also two of the three terms in the sum are symmetric -- if the two exponential rates were equal, their expectations would cancel out, but in any case computing one expectation will also tell you the form of the other).
