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


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?


  • 2
    $\begingroup$ This clearly belongs on stats.SE. The end application does not matter when the question is this technical. $\endgroup$
    – kurtosis
    Commented Aug 15, 2020 at 18:11

2 Answers 2


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:

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


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

(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).

  • $\begingroup$ Agree with your expressions, see also my last comment below. Giving you both +10, thanks. $\endgroup$
    – user34971
    Commented Aug 16, 2020 at 5:54
  • 1
    $\begingroup$ One more variable change shows we get back to square 1. I added a note. $\endgroup$
    – ir7
    Commented Aug 16, 2020 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.

  • $\begingroup$ Can you elaborate? $\endgroup$
    – user34971
    Commented Aug 15, 2020 at 20:18
  • $\begingroup$ Firstly, fundamental theorem of calculus does not apply to improper integral. Secondly $q$ itself may not even be continuous let alone differentiable. $\endgroup$
    – CABLE
    Commented Aug 15, 2020 at 20:21
  • $\begingroup$ btw, even if you assume $q$ is very nice, you would have a $q(\infty)$ on the right hand side $\endgroup$
    – CABLE
    Commented Aug 15, 2020 at 20:47
  • $\begingroup$ 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. $\endgroup$
    – user34971
    Commented Aug 15, 2020 at 20:48
  • 1
    $\begingroup$ 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. $\endgroup$
    – user34971
    Commented Aug 16, 2020 at 5:53

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