<p>Hint: Simply apply integration by parts to the integral on the LHS. Simplify and you should arrive at the following expression:
<span class="math-container">\begin{equation}
(1-R)-\int_{R-k}^1G(\theta)\mathrm d\theta.
\end{equation}</span>
Add and subtract <span class="math-container">$k$</span> to obtain:
<span class="math-container">\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}</span>
Observe that <span class="math-container">$(1-(R-k))=\int_{R-k}^1(1)\mathrm d\theta$</span>. Substitute that in and collect terms, you'll get the RHS expression.</p>

CC BY-SA 4.0
2021-01-29T03:28:06.590Z
2021-01-29T03:28:06.590Z
Herr K.
<p>I am told that the following equality follows from integration by parts:</p>
<p><span class="math-container">$$\int_{R-k}^{1}(\theta-R)dG(\theta)-G(R-k)k=\int_{R-k}^{1}(1-G(\theta))d\theta-k$$</span> Where <span class="math-container">$R&gt;k&gt;0$</span> and <span class="math-container">$G$</span> is the CDF of <span class="math-container">$\theta$</span> which is distributed on <span class="math-container">$[0,1]$</span>. Can someone explain how integration by parts has been used here? Thank you.</p>

<microeconomics><mathematical-economics><probability>
Integration by parts with CDF
Charles
# Integration by parts with CDF

(Asked by: username_0 on 2021-01-28)

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 $R>k>0$ 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.

**Tags:** microeconomics, mathematical-economics, probability

---

## 1 answers

### Answer by username_1 on 2021-01-29. Score: 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{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.


---

