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

• This clearly belongs on stats.SE. The end application does not matter when the question is this technical. – kurtosis Aug 15 '20 at 18:11

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).

• Agree with your expressions, see also my last comment below. Giving you both +10, thanks. – ilovevolatility Aug 16 '20 at 5:54
• One more variable change shows we get back to square 1. I added a note. – ir7 Aug 16 '20 at 6:06

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.

• Can you elaborate? – ilovevolatility Aug 15 '20 at 20:18
• Firstly, fundamental theorem of calculus does not apply to improper integral. Secondly $q$ itself may not even be continuous let alone differentiable. – CABLE Aug 15 '20 at 20:21
• btw, even if you assume $q$ is very nice, you would have a $q(\infty)$ on the right hand side – CABLE Aug 15 '20 at 20:47
• FTC can be extended to improper integrals. I will think more about your point regarding differentiability, but I am not sure (yet) this is the answer. EDIT: well q will go to zero as x goes to infinity. – ilovevolatility Aug 15 '20 at 20:48
• Yes, I think you are right that $q = q(a,y(a))$ which if not taken into account leads to garbage out. And/or, using @ir7 notation, $Y(a)$ is a different random variable for each value of the parameter $a$, hence I don't know if $dE[Y(a)]/da$ is even well-defined. – ilovevolatility Aug 16 '20 at 5:53