# Expected value of $\dfrac 1 {I(y_1<c) + I(y_2<c)}$, where $y_1$ and $y_2$ are i.i.d. random variables with exponential distribution

There is a random variable $r = \dfrac 1 {I(y_1<c) + I(y_2<c)}$. Both $y_1$ and $y_2$ are i.i.d. random variables with exponential distribution, so their joint distribution is $f(y_1, y_2) = \prod_{i=1}^2 \theta^{-1} e^{- \dfrac {y_i} \theta}$.

My goal is to calculate expected value $E(r)$. I tried it myself and got stuck at some point. Below $I(y<c)$ is the indicator function. This is what I have attempted so far :

\begin{align} E(r) &= \int_0^{\infty} \int_0^{\infty} \dfrac 1 {I(y_1<c) + I(y_2<c)} \theta^{-1} e^{- \dfrac {y_1} \theta} \theta^{-1} e^{- \dfrac {y_2} \theta} dy_1 dy_2 \\ &= \int_0^{\infty} \bigg( \int_0^{c} \dfrac 1 {1 + I(y_2<c)} \theta^{-1} e^{- \dfrac {y_1} \theta} dy_1 + \int_c^{\infty} \dfrac 1 {I(y_2<c)} \theta^{-1} e^{- \dfrac {y_1} \theta} dy_1 \bigg) \ \theta^{-1} e^{- \dfrac {y_2} \theta} dy_2 \\ &= F_{y_1}(c) \int_0^{\infty} \dfrac 1 {1 + I(y_2<c)} \theta^{-1} e^{- \dfrac {y_2} \theta} dy_2 \ + (1-F_{y_1}(c)) \int_0^{\infty} \dfrac 1 {I(y_2<c)} \theta^{-1} e^{- \dfrac {y_2} \theta} dy_2\\ &= F_{y_1}(c) \bigg(\dfrac 1 2 F_{y_2}(c) + (1-F_{y_2}(c)) \bigg) \ + (1-F_{y_1}(c) ) \bigg(F_{y_2}(c) + \int_c^{\infty} \dfrac 1 {I(y_2<c)} \theta^{-1} e^{- \dfrac {y_2} \theta} dy_2 \bigg) \end{align}

I am stuck because $I(y_2<c) = 0$ in that integral. How should I solve this problem?

This random variable has a positive probability to be infinite, therefore its expectation is $+\infty$.