I am stucked on this integral, which is from a research paper in Finance, for a while, so can anyone please help walk me through how we can get the answer on the RHS of this integral?

Prove: $\int_{-\infty}^{\infty} \max\ (0,\frac{t\hat{a}+sy}{t+s})\sqrt{\frac{H}{2\pi}}e^{-\frac{H}{2}(y-\hat{a})^{2}}dy= \hat{a}\ N[-(\frac{-t\hat{a}}{s}-\hat{a})\sqrt{H}] +\frac{\sqrt{H}}{t}\phi [(\frac{-t\hat{a}}{s}-\hat{a})\sqrt{H}]$

where N is distribution function of standard normal random variable, $\phi$ is its corresponding density function, and y follows normal distribution $N(\hat{a}, \frac{1}{s})$.

I also can't see how from the equation above, the author came up with these partial derivatives, given $H=\frac{st}{s+t}$:

$\frac{\partial\ V}{\partial\hat{a}} = N + \hat{a}\phi\sqrt{H} + \frac{H}{t}(\frac{-t\hat{a}}{s}-\hat{a})\phi\sqrt{H}$ = N, $\ \frac{\partial\ V}{\partial\frac{t\hat{a}}{s}}=0$

$\frac{\partial V}{\partial t} = (-1+\frac{s}{2s+2t})(\frac{\sqrt{H}}{t^2})\phi$, $\frac{\partial V}{\partial s} = \frac{t}{2(t+s)^2\sqrt{H}}$


Hint: For $\beta>0$, we have: $$\int_{-\infty}^\infty \max(0,\alpha+\beta x)\phi(x)\,dx = \int_{-\alpha/\beta}^\infty (\alpha+\beta x)\phi(x)\,dx =\alpha N(\alpha/\beta) + \beta\phi(\alpha/\beta), $$ by using $$ \int \phi(x)dx = N(x) +C, \; \int x\phi(x)dx = -\phi(x) +C. $$

To use it in your calculation apply change of variable $x = \sqrt{H}(y-\hat{a})$.

Edit: It seems to me that original random variable should have been ${\rm Normal}(\hat{a}, 1/H)$ (not ${\rm Normal}(\hat{a}, 1/s)$).

Edit2: Denoting by $V(\alpha, \beta):= \alpha N(\alpha/\beta) + \beta\phi(\alpha/\beta)$, we get $$\frac{\partial V}{\partial \alpha} = N(\alpha/\beta)+ \alpha/\beta N'(\alpha/\beta) +\phi'(\alpha/\beta) $$ $$ = N(\alpha/\beta)+ \alpha/\beta \phi(\alpha/\beta) -\alpha/\beta\phi(\alpha/\beta) $$ $$ = N(\alpha/\beta).$$

Edit3: $$ \int_{-\infty}^\infty\max\left(0,\frac{t\hat{a}+sy}{t+s}\right)\sqrt{\frac{H}{2\pi}}e^{-\frac{H}{2}(y-\hat{a})^{2}}dy$$ $$= \int_{-\infty}^\infty\max \left(0,\frac{t\hat{a}+s\left(x/\sqrt{H}+\hat{a}\right)}{t+s}\right)\sqrt{\frac{1}{2\pi}}e^{-\frac{1}{2}x^{2}}dx $$ $$= \int_{-\infty}^\infty\max \left(0,\hat{a}+\frac{s}{\sqrt{H}(t+s)}x \right)\phi(x) dx $$ We now take $$\alpha = \hat{a}$$ and $$ \beta = \frac{s}{\sqrt{H}(t+s)}, \quad \frac{\alpha}{\beta} = \left(\hat{a}+\frac{t\hat{a}}{s}\right)\sqrt{H}. $$ Using the formula above, we get: $$ \hat{a} N\left(\left(\hat{a}+\frac{t\hat{a}}{s}\right)\sqrt{H}\right) + \frac{s}{\sqrt{H}(t+s)}\phi\left(\left(\hat{a}+\frac{t\hat{a}}{s}\right)\sqrt{H}\right)$$ Using $H:=st/(s+t)$ (hence $ \beta = \sqrt{H}/t$) as in the source paper, we have further: $$ V(\hat{a},s,t):=\hat{a} N\left(\left(\hat{a}+\frac{t\hat{a}}{s}\right)\sqrt{H}\right) + \frac{\sqrt{H}}{t}\phi\left(\left(\hat{a}+\frac{t\hat{a}}{s}\right)\sqrt{H}\right) $$

Edit4 Similar to Edit2: $$\frac{\partial V}{\partial\hat{a}}= N\left(\left(\hat{a}+\frac{t\hat{a}}{s}\right)\sqrt{H}\right). $$

Edit5 Introducing variable $Y:=\hat{a}t/s$, we have $$ V= \hat{a} N\left(\left(\hat{a}+Y\right)\sqrt{H}\right) + \frac{\sqrt{H}}{t}\phi\left(\left(\hat{a}+Y\right)\sqrt{H}\right). $$ So: $$\frac{\partial V}{\partial Y} = \hat{a} \phi\left(\left(\hat{a}+Y\right)\sqrt{H}\right) \sqrt{H} -\frac{\sqrt{H}}{t}\left(\hat{a}+Y\right)\sqrt{H}\phi\left(\left(\hat{a}+Y\right)\sqrt{H}\right)\sqrt{H} = 0,$$ due to observation (using definition of $Y$ and $H$): $$ \frac{\sqrt{H}}{t}\left(\hat{a}+Y\right)\sqrt{H} = \hat{a}.$$

Edit6 Noting that:

$$V= \hat{a} N\left(\frac{\hat{a}t}{\sqrt{H}}\right) + \frac{\sqrt{H}}{t}\phi\left(\frac{\hat{a}t}{\sqrt{H}}\right),$$

we get:

$$\frac{\partial V}{\partial s} =\hat{a} \phi\left(\frac{\hat{a}t}{\sqrt{H}}\right) \left(\frac{\hat{a}t}{\sqrt{H}}\right)'_s +\left( \frac{\sqrt{H}}{t}\right)'_s\phi\left(\frac{\hat{a}t}{\sqrt{H}}\right) $$ $$ - \frac{\sqrt{H}}{t} \frac{\hat{a}t}{\sqrt{H}} \phi\left(\frac{\hat{a}t}{\sqrt{H}}\right)\left(\frac{\hat{a}t}{\sqrt{H}}\right)'_s$$ $$ = \left( \frac{\sqrt{H}}{t}\right)'_s\phi\left(\frac{\hat{a}t}{\sqrt{H}}\right)=\frac{H'_s}{2t\sqrt{H}} \phi\left(\frac{\hat{a}t}{\sqrt{H}}\right)= \frac{t}{2(t+s)^2\sqrt{H}}\phi\left(\frac{\hat{a}t}{\sqrt{H}}\right)$$

Edit7: Same calculation sequence gives:

$$\frac{\partial V}{\partial t} = \left( \frac{\sqrt{H}}{t}\right)'_t\phi\left(\frac{\hat{a}t}{\sqrt{H}}\right)=\frac{\frac{H'_t}{2\sqrt{H}}t-\sqrt{H}}{t^2}\phi\left(\frac{\hat{a}t}{\sqrt{H}}\right)$$ $$ = \left(\frac{H'_t}{2{H}}t -1\right) \frac{\sqrt{H}}{t^2}\phi\left(\frac{\hat{a}t}{\sqrt{H}}\right)=\left( \frac{s}{2(s+t)}-1\right) \frac{\sqrt{H}}{t^2}\phi\left(\frac{\hat{a}t}{\sqrt{H}}\right)$$

  • $\begingroup$ Thank you very much for your helpful answer. May you elaborate on how did you get the two integrals, and how the two integrals become when $x\rightarrow \infty$? And do you see how we can end up with the two partial derivatives above from the equation? $\endgroup$ – ghjk Dec 21 '14 at 2:54
  • $\begingroup$ Btw, even after using your two integrals and the suggested change of variables, I don't see how the coefficients $\frac{1}{t+s}$ and $\frac{s}{t+s}$ became $\hat{a}$ and $\frac{\sqrt{H}}{t}$, respectively. $\endgroup$ – ghjk Dec 21 '14 at 3:26
  • $\begingroup$ $N'(x)=\phi(x)$ and $\phi'(x)=-x\phi(x)$ by their definitions as functions, and $\phi(x)$ goes to $0$ when $x$ goes to $\infty$. $\endgroup$ – ir7 Dec 21 '14 at 3:29
  • $\begingroup$ I added Edit2 as an example of manipulation of derivatives with respect to parameters. $\endgroup$ – ir7 Dec 21 '14 at 3:30
  • 1
    $\begingroup$ I also got the same for that derivative! Many thanks for your great help, ir7:) Merry Xmas and Happy New Year to you and your family!! $\endgroup$ – ghjk Dec 24 '14 at 3:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.