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I am trying to calculate the expected benefit of an action, where both the benefit and probability of carrying out the action depends on a parameter c, which is distributed according to f, F'=f.

The benefit of the action is given by $\delta b_h-c$, and therefore the action will only be carried out if $c \le \delta b_h$.

The probability of that is $\int_{-\infty}^{\delta b_h} f dc = F(\delta b_h)$.

But since the benefit does also depend of the draw of c, i am not sure how to calculate the expected value.

Any help is greatly appreciated.

Best, Esben

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Lat $a=\delta b_h$. The benefit is $$\mathbb E((a-c)^+)=\displaystyle\int_{-\infty}^a(a-x)f_c(x)\mathrm dx=aF(a)-\int_{-\infty}^axf_c(x)\mathrm dx. $$ Edit: If $c$ is standard normal, $\mathbb E((a-c)^+)=a\Phi(a)+\varphi(a)$, where $$ \varphi(a)=\frac1{\sqrt{2\pi}}\mathrm e^{-a^2/2},\qquad\Phi(a)=\int_{-\infty}^a\varphi(x)\mathrm dx. $$ Edit-edit: Since $x\varphi(x)=-\varphi'(x)$ and $\varphi(-\infty)=0$, $$ \int_{-\infty}^{a}x\varphi(x)\mathrm dx=\int_{-\infty}^{a}-\varphi'(x)\mathrm dx=\left[-\varphi(x)\right]_{-\infty}^{a}=-\varphi(a). $$

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Great. thanks! But is the result of the last integral? The one you subtract. –  E_T Jan 11 '13 at 12:43
    
No simpler general expression (but perhaps you have a specific distribution of $c$ in mind?). –  Did Jan 11 '13 at 17:26
    
According to a normal distribution? –  E_T Jan 12 '13 at 12:20
    
See Edit. $ $ $ $ –  Did Jan 12 '13 at 12:31
    
Will you take me through the steps? It'd be GREAT! I guess this is how you get to the general solution? $\int_{-\infty}^{\delta b_h}\delta b_h f_c(x) -\int_{-\infty}^{\delta b_h}xf_c(x) =\delta b_hF(\delta b_h)-\int_{-\infty}^{\delta b_h}xf_c(x)\mathrm dx$. But how do you get from $-\int_{-\infty}^{\delta b_h}xf_c(x)\mathrm dx$ to $\frac1{\sqrt{2\pi}}\mathrm e^{-(\delta b_h)^2/2}$ (I know its the formula for the dist) –  E_T Jan 12 '13 at 13:34

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