Differentiating an Expected Square Root with a Parameter-Dependent Density

# $\frac{\partial}{\partial a} E [\sqrt{a+X} ]$, $X > 0$ a.s., $a \geq 0$


# $\frac{\partial}{\partial a} E [\sqrt{a+X} ]$, $X > 0$ a.s., $a \geq 0$












Although maybe this could have been posted at cross-validated, I actually have a financial application in mind.

Problem:

There is a very elementary mistake somewhere, but I can't see it:

Let $X$ be a random variable with $X > 0$ almost surely. Let $a$ be a non-negative real-number. Denote by $p(x)$ the probability density of $X$. Then, $$ E [\sqrt{a + X} \;] = \int_0^\infty \sqrt{a+x}\; p(x) dx $$ and $$ \frac{\partial}{\partial a} E [\sqrt{a+X}\; ] = \frac{1}{2}\int_0^\infty \frac{1}{\sqrt{a+x}}\; p(x) dx > 0 $$

On the other hand, we can also consider the probability density $q(\sqrt{a+x})$ of the random variable $\sqrt{a+X}$ directly, and since $\sqrt{a+X} > \sqrt{a}$ almost surely, $$ E [\sqrt{a + X} \;] = \int_\sqrt{a}^\infty \sqrt{a+x}\; q(\sqrt{a+x}) d\sqrt{a+x} $$ Now, $$ d\sqrt{a+x} = \frac{1}{2} \frac{dx}{\sqrt{a+x}} $$ and hence $$ E [\sqrt{a + X} \;] = \frac{1}{2} \int_0^\infty q(\sqrt{a+x}) dx $$ Differentiate the above expressiont wrt to $a$: \begin{align} \frac{\partial}{\partial a} E [\sqrt{a + X} \;] &= \frac{1}{2} \int_0^\infty \frac{1}{2\sqrt{a+x}} \frac{\partial q(\sqrt{a+x})}{\partial \sqrt{a+x}} dx\\ &= \frac{1}{2} \int_\sqrt{a}^\infty \frac{\partial q(\sqrt{a+x})}{\partial \sqrt{a+x}} d\sqrt{a+x} \\ &= - \frac{1}{2} q(\sqrt{a}) \end{align}

Since 1. the sign is wrong, and 2. $q(\sqrt{a}) = 0$, this is (twice) in contradiction with what was derived earlier, namely, $$ \frac{\partial}{\partial a} E [\sqrt{a+X}\; ] = \frac{1}{2}\int_0^\infty \frac{1}{\sqrt{a+x}}\; p(x) dx > 0 $$

So where did I go wrong?

Thanks.

## Answer by ir7 (score 3, accepted)

https://quant.stackexchange.com/a/57375

I'm not sure what your $q$ is (it doesn't seem well defined). For clarity, let $$ Y = \sqrt{a+X} > \sqrt a \; \; a.s. $$

For cdf's we have: $$ F_Y(y) = P(Y\leq y) = P(\sqrt{a+X}\leq y) = P(X\leq y^2-a)=F_X(y^2-a) $$

By taking derivatives, we get the following relationship between pdf's:

$$ p_Y(y) = 2y p_X(y^2-a) $$

So:

$$E[Y] = \int_{\sqrt{a}}^\infty y p_Y(y) dy =\int_{\sqrt{a}}^\infty 2y^2 p_X(y^2-a) dy $$ $$= \int_{{0}}^\infty \sqrt{a+x}p_X(x) dx = E[\sqrt{a+X}]$$

(after a variable transformation $x=y^2-a$ in the third equality).

We then may want to take the derivative wrt to $a$ of: $$E[Y] = \int_{\sqrt{a}}^\infty y p_Y(y) dy = \int_{{0}}^\infty 2^{-1}p_Y(\sqrt{z+a}) dz $$

(after transformation $y=\sqrt{z+a}$), which brings us back to square 1 (given the pdf relationships).

## Answer by CABLE (score 2)

https://quant.stackexchange.com/a/57370

Your last claim $\frac{1}{2} \int_{\sqrt{a}}^\infty \frac{\partial q(\sqrt{a+x})}{\partial \sqrt{a + x}} d \sqrt{a+x} = -\frac{1}{2}q(\sqrt{a})$ is not true.

Realized that the part above is irrelevant. Assuming $q$ is nice enough, the problems lies in the part of taking derivative $\frac{\partial q(\sqrt{a+x})}{\partial a}$. The mistake is that $q_a(y) = q(a, y)$ itself is also a function of $a$. So when taking derivative, we need take care of both arguments.