有关对称函数

opuikl_0

考虑如下空间:
\[
\mathcal{S}:=\{S \text { on }(\Omega, \mathcal{F}, \mathbb{P}): S>0 \text { almost surely and } \mathbb{E}(S)=1\} .
\]
对任意这个空间里的$S$, 定义它的对称版本$\hat S$:
\[
\mathbb{P}(\hat{S} \leq z)=\mathbb{E}\left(S \mathbb{1}_{\{S \geq 1 / z\}}\right), \quad \text { for any } z>0 .
\]
（1）证明如下关系式：(这是答案，我不能理解第一行的积分是怎么来的？）
\[
\begin{aligned}
\mathbb{E}^{\hat{S}^{1-u}} & =\int_{\mathbb{R}_{+}}(1-u) x^{-u} \mathbb{P}(\hat{S}>x) \mathrm{d} x \\
& =\int_{\mathbb{R}_{+}}(1-u) x^{-u} \mathbb{E}\left(S 1_{\{S<1 / x\}}\right) \mathrm{d} x \\
& =\mathbb{E}\left(\int_0^{1 / S}(1-u) x^{-u} S \mathrm{~d} x\right)=\mathbb{E}\left(S^u\right)
\end{aligned}
\]（2）已知下面的两个关系式：
\[
\begin{aligned}
& C_{\mathrm{BS}}(-x, t, \sigma)=1-\mathrm{e}^{-x}+\mathrm{e}^{-x} C_{\mathrm{BS}}(x, t, \sigma) \\
& 1-\mathrm{e}^{-x}+\mathrm{e}^{-x} \mathbb{E}\left(S-\mathrm{e}^x\right)_{+}=\mathbb{E}\left[S\left(S^{-1}-\mathrm{e}^{-x}\right)_{+}\right]
\end{aligned}
\]要求证明下面的关系式：（这是答案，我看不懂最后一行。。）
\begin{aligned}
C_{\mathrm{BS}}\left(-x, t, \sigma_t(x)\right) & =1-\mathrm{e}^{-x}+\mathrm{e}^{-x} C_{\mathrm{BS}}\left(x, t, \sigma_t(x)\right) \\
& =1-\mathrm{e}^{-x}+\mathrm{e}^{-x} \mathbb{E}\left(S_t-\mathrm{e}^x\right)_{+} \\
& =\mathbb{E}\left[S_t\left(S_t^{-1}-\mathrm{e}^{-x}\right)_{+}\right] \\
& =C_{\mathrm{BS}}\left(-x, t, \hat{\sigma}_t(-x)\right) .
\end{aligned}

战巡

第一问是个很常用的结论：
令$\phi(x)$为$R^+$上连续的单调函数，$X>0$  $a.s.$，有
\[E(\phi(X))=\phi(0)+\int_0^{+\infty}(1-F(x))d\phi(x)\]
证明并不困难
\[\int_0^{+\infty}(1-F(x))d\phi(x)=\int_0^{+\infty}\int_x^{+\infty}dF(y)d\phi(x)\]
易证这玩意可积，由Tonelli定理知可交换积分次序
\[=\int_0^{+\infty}\int_0^yd\phi(x)dF(y)=\int_0^{+\infty}(\phi(x)-\phi(0))dF(y)=E(\phi(x))-E(\phi(0))=E(\phi(x))-\phi(0)\]

至于第二问我不知道是什么鬼东西，不同的书用不同的符号，你最好解释清楚再问

opuikl_0

回复 2# 战巡

谢谢！看明白了！

hbghlyj

战巡 发表于 2017-5-1 03:37
第一问是个很常用的结论：
令$\phi(x)$为$R^+$上连续的单调函数，$X>0$  $a.s.$，有
\[E(\phi(X))=\phi(0)+\int_0^{+\infty}(1-F(x))d\phi(x)\]
证明并不困难
\[\int_0^{+\infty}(1-F(x))d\phi(x)=\int_0^{+\infty}\int_x^{+\infty}dF(y)d\phi(x)\]
易证这玩意可积，由Tonelli定理知可交换积分次序
\[=\int_0^{+\infty}\int_0^yd\phi(x)dF(y)=\int_0^{+\infty}(\phi(x)-\phi(0))dF(y)=E(\phi(x))-E(\phi(0))=E(\phi(x))-\phi(0)\]

Unclear detail in an application of Fubini's theorem
Let $X$ be a positive r.v. and $f: \mathbb{R}^+ \rightarrow \mathbb{R}$ be a differentiable function with a continuous derivative such that $f(X)$ is integrable. Show that $$\Bbb E[f(X)] = f(0) + \int_0^{+\infty} f'(t)P(X\geq t)\,dt.$$
Proof of an alternative form of expectation of a function of a random variable.
Let X be a positive r.v. and $f : \mathbb{R}^{+} \rightarrow \mathbb{R}$ a differentiable function
with continuous derivative and such that $f(X)$ is integrable. Then
$\mathrm{E}[f(X)]=f(0)+\int_{0}^{+\infty} f^{\prime}(t) \mathrm{P}(X \geq t) d t$
Expected value of a function in terms of CDF