<p>Your first question was addressed in another answer. For the second question (why $\Phi(b) \ne 1$), you are confusing the cdf and the density. It is true that $\int_{a}^{b}{f(t)dt}=1$ but not that $\int_{a}^{b}{F(t)dt}=1$. Think for instance of $[a,b]=[0,1]$ and the uniform distribution ($f(t)=1,F(t)=t$ for all $t \in [0,1]$).</p>

CC BY-SA 4.0
2018-05-04T19:12:15.737Z
2018-05-04T19:12:15.737Z
Oliv
<p>That's just a question of notation. Note that</p>

<p>$$\mathcal{d}F(t) = f(t)\mathcal{d}t$$</p>

<p>Then the proof should be easy to understand.</p>

CC BY-SA 4.0
2018-05-04T10:46:10.083Z
2018-05-04T10:46:10.083Z
saguru
<p>$\int_a^b u(x)dF(x)$ (1)$ = u(t)F(t)|_a^b - \int_a^b F(t)u^\prime(t)dt$ (2)$ = u(b)-\int_a^b F(t)u^\prime(t)dt$</p>

<p>$= u(b)-(\Phi(t)u^\prime(t)|_a^b-\int_a^b \Phi(t)u^{\prime\prime}(t)dt=u(b)-\Phi(b)u^\prime(b)+\int_a^b \Phi(t)u^{\prime\prime}(t)dt$.  </p>

<p>From (1) to (2), I know that it is a partial derivative, but shoudn't it be $u(t)F(t)|_a^b - \int_a^b F(t)u^\prime(t)dF(t)$?? How can we transform $dF(t)$ to $dt$?</p>

<p>I also want to know why $\Phi(b)$ is not equal to 1. </p>

<p>I attach the original paper in case the information is not enough. <a href="https://i.sstatic.net/QCWkQ.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/QCWkQ.png" alt="enter image description here"></a></p>

<microeconomics><theory><probability>
understanding the proof of stochastic dominance.
shk910
# understanding the proof of stochastic dominance.

(Asked by: username_0 on 2018-05-04)

$\int\_a^b u(x)dF(x)$ (1)$ = u(t)F(t)|\_a^b - \int\_a^b F(t)u^\prime(t)dt$ (2)$ = u(b)-\int\_a^b F(t)u^\prime(t)dt$

$= u(b)-(\Phi(t)u^\prime(t)|\_a^b-\int\_a^b \Phi(t)u^{\prime\prime}(t)dt=u(b)-\Phi(b)u^\prime(b)+\int\_a^b \Phi(t)u^{\prime\prime}(t)dt$.

From (1) to (2), I know that it is a partial derivative, but shoudn't it be $u(t)F(t)|\_a^b - \int\_a^b F(t)u^\prime(t)dF(t)$?? How can we transform $dF(t)$ to $dt$?

I also want to know why $\Phi(b)$ is not equal to 1.

I attach the original paper in case the information is not enough. [![enter image description here](https://i.sstatic.net/QCWkQ.png)](https://i.sstatic.net/QCWkQ.png)

**Tags:** microeconomics, theory, probability

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## 2 answers

### Answer by username_2 on 2018-05-04. Score: 5 | Accepted answer

That's just a question of notation. Note that

$$\mathcal{d}F(t) = f(t)\mathcal{d}t$$

Then the proof should be easy to understand.


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### Answer by username_1 on 2018-05-04. Score: 2

Your first question was addressed in another answer. For the second question (why $\Phi(b) \ne 1$), you are confusing the cdf and the density. It is true that $\int\_{a}^{b}{f(t)dt}=1$ but not that $\int\_{a}^{b}{F(t)dt}=1$. Think for instance of $[a,b]=[0,1]$ and the uniform distribution ($f(t)=1,F(t)=t$ for all $t \in [0,1]$).


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