New answers tagged probability
3
Hint: Simply apply integration by parts to the integral on the LHS. Simplify and you should arrive at the following expression:
\begin{equation}
(1-R)-\int_{R-k}^1G(\theta)\mathrm d\theta.
\end{equation}
Add and subtract $k$ to obtain:
\begin{equation}
(1-R+k-k)-\int_{R-k}^1G(\theta)\mathrm d\theta = (1-(R-k))-k-\int_{R-k}^1G(\theta)\mathrm d\theta.
\end{...
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