A conditional expected value

Question

How can you easily calculate $$\mathbb{E}\left(\max\left(0,W_{1}\right)\mid\mathcal{F}_{t}\right)\;\;\;t\in\left[0,1\right],$$ where $W$ denotes a Wiener process in its natural filtration: $\mathcal{F}$?

I did this so far, but I can't continue...

$$\mathbb{E}\left(\max\left(0,W_{1}\right)\mid\mathcal{F}_{t}\right)=\mathbb{E}\left(\max\left(0,W_{1}-W_{t}+W_{t}\right)\mid\mathcal{F}_{t}\right)=...$$ where $W_{1}-W_{t}$ is independent from $\mathcal{F}_{t}$, however $W_{t}$ is $\mathcal{F}_{t}$ measurable. So $$...=\left.\mathbb{E}\left(\max\left(0,W_{1}-W_{t}+x\right)\right)\right|_{x=W_{t}},$$ but I can't calculate $\mathbb{E}\left(\max\left(0,W_{1}-W_{t}+x\right)\right)$. $$\begin{aligned} \mathbb{E}\left(\max\left(0,W_{1}-W_{t}+x\right)\right) & =\mathbb{E}\left(\max\left(0,N\left(x,1-t\right)\right)\right)= \\&=\int_{-\infty}^{\infty}\max\left(0,y\right)\cdot\frac{1}{\sqrt{2\pi\left(1-t\right)}}\exp\left\{ -\frac{\left(y-x\right)^{2}}{2\left(1-t\right)}\right\} dy=\\ &=\int_{0}^{\infty}y\cdot\frac{1}{\sqrt{2\pi\left(1-t\right)}}\exp\left\{ -\frac{\left(y-x\right)^{2}}{2\left(1-t\right)}\right\} dy=? \end{aligned}$$ Unfortunatelly I can't calculate the last integral. Is there any tip? I have seen this page: First approximation of the expected value of the positive part of a random variable, but to be honest it didn't really help.

Answer 1

Define $\phi$ as the standard normal pdf. We have $$\begin{aligned}E[(W_T)^+|W_t=x] &=\int_{\mathbb{R}}(w)^+\frac{1}{\sqrt{2\pi (T-t)}}\exp\bigg\{-\frac{(w-x)^2}{2(T-t)}\bigg\}dw= \\ &= \int_{[0,\infty)}w\frac{1}{\sqrt{2\pi (T-t)}}\exp\bigg\{-\frac{(w-x)^2}{2(T-t)}\bigg\}dw= \\ &= \int_{[0,\infty)}\frac{w}{\sqrt{T-t}}\,\phi\bigg(\frac{w-x}{\sqrt{T-t}}\bigg)dw= \\ &= \sqrt{T-t}\int_{[0,\infty)}\frac{x+(w-x)}{T-t}\,\phi\bigg(\frac{w-x}{\sqrt{T-t}}\bigg)dw= \\ &= \sqrt{T-t}\bigg(\int_{[0,\infty)}\frac{x}{T-t}\,\phi\bigg(\frac{w-x}{\sqrt{T-t}}\bigg)dw-\int_{[0,\infty)}\frac{\partial}{\partial w}\phi\bigg(\frac{w-x}{\sqrt{T-t}}\bigg)dw\bigg)= \\ &=x\,P_{\mathcal{N}(x,T-t)}(w\geq 0)+\sqrt{T-t}\,\phi \bigg(\frac{x}{\sqrt{T-t}}\bigg) \end{aligned}$$ where I have used that $\phi'(x)=-x\phi(x)$. In your case, $T=1$.

Answer 2

Hint: just introduce a change of variable $w = y - x$ and the integral separates into two pieces, one being a simple exponential, the other just the normal CDF.