Integration by parts with CDF I am told that the following equality follows from integration by parts:
$$\int_{R-k}^{1}(\theta-R)dG(\theta)-G(R-k)k=\int_{R-k}^{1}(1-G(\theta))d\theta-k$$ Where $R&gt;k&gt;0$ and $G$ is the CDF of $\theta$ which is distributed on $[0,1]$. Can someone explain how integration by parts has been used here? Thank you.
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{equation}
Observe that $(1-(R-k))=\int_{R-k}^1(1)\mathrm d\theta$. Substitute that in and collect terms, you'll get the RHS expression.