Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a fairly basic question that relates to understanding a particular derivation.

I have the following function $Q(x) = E\left[I(F(x+\varepsilon)>c)\right]$, where $x \in R$, $\varepsilon \sim N(0,\sigma^2)$ for a known $\sigma>0$, $c$ is a scalar constant, $F(\cdot)$ is a known, bounded function, and $I(\cdot)$ is an indicator function.

This can be written as:

$E[I(F(x+\varepsilon)>c)] = \frac{1}{{\sqrt{2\pi}}\sigma} \int{I(F(x+\varepsilon)>c) \exp\left(-\frac{\varepsilon^2}{2\sigma^2} \right)}d\varepsilon $

Let $y = x + \varepsilon$ and rewrite the above as:

$E[I(F(x+\varepsilon)>c)] = \frac{1}{{\sqrt{2\pi}}\sigma} \int{I(F(y)>c) \exp\left(-\frac{(y-x)^2}{2\sigma^2} \right)}dy$

I am interested in calculating $\frac{\partial Q(x)}{\partial x}$. The derivative is:

$\frac{\partial E[I(F(x+\varepsilon)>c)]}{\partial x}= \frac{1}{{\sqrt{2\pi}}\sigma} \int{I(F(y)>c) \exp\left(-\frac{(y-x)^2}{2\sigma^2} \right)} \frac{(y-x)}{\sigma^2}dy \quad \quad $ (1)

Substituting back in again and rearranging slightly:

$\frac{\partial E[I(F(x+\varepsilon)>c)]}{\partial x} = \frac{1}{\sigma^2} E[I(F(x+\varepsilon)>c)\varepsilon]$

Finally, here are my two questions: (1) is the above derivation correct? (2) If so, I am trying to understand why in eq(1) I do not need to take the derivative of $I(F(y)>c)$ with respect to $x$. I know there has been a change of variables, but $y$ is obviously a function of $x$. Yet I was told that "once you do the substitution, $y=x+\varepsilon$, then y is just a dummy variable (i.e. an index) which runs from -infinity to infinity, and one cannot take the derivative in a dummy variable." I know you can't take the derivative of an indicator function, but somehow I can't get this clear in my head.

Thanks.

More broadly, in my actual problem, $x$ and $\varepsilon$ are high dimensional and I am trying to compute the derivative via Monte Carlo integration....

share|improve this question
add comment

1 Answer

up vote 0 down vote accepted

This is true and might be recognized as the derivative of the function $x\mapsto\mathrm P(x+\varepsilon\in A)$, where $A=\{F\gt c\}$. Thus, one is considering $U(x)=\mathrm E(u(x+\varepsilon))$, for $u=\mathbf 1_A$ and the assertion is that $U'(x)=\sigma^{-2}\mathrm E(\varepsilon u(x+\varepsilon))$. This holds in full generality and the proof is as you explain.

Re your (2), the derivative of $\mathbf 1_{F(y)\gt c}$ with respect to $x$ exists and it is zero, simply because the function does not depend on $x$. Recall that, under suitable hypotheses, $$ \frac{\mathrm d}{\mathrm dx}\int f(x,y)\,\mathrm dy=\int \frac{\partial f(x,y)}{\partial x}\,\mathrm dy. $$

share|improve this answer
    
did: Thanks for confirming that the derivation is correct. On the the second point, why should I not think of the expression as $1_{F(y(x,\varepsilon))>c}$, which is a function of $x$? Further, if the derivative w.r.t x exists and it is zero, then wouldn't this zero appear inside the integrand in my eq(1)? I would greatly appreciate it if you could elaborate more, as I'm really hoping to make the logic crystal clear for myself. Thanks again for the help so far. –  user36363 Jul 23 '12 at 18:43
    
There is no $y(x,\varepsilon)$ here. Nowhere is $y$ a random variable, $y$ is a variable of integration and could be replaced by any symbol you like. –  Did Jul 23 '12 at 22:18
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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