# Thread: Maximizing over a continuum of possibilities

1. ## Maximizing over a continuum of possibilities

Let x be a random variable with density function f and [a,b] be the possible set of values of x.

Let w(x) be a value associated with each value of x. I.e. w() is not known, and suppose that U(w(x)) is a $C^{\infty}$ function.

Let $H(w(x)) = \int_{a}^b f(x)U(w(x))dx$

Find the derivative of H with respect to w over all x.

I don't know where to begin because I'm not even sure I understand the question.

I suspect the answer is f(x)U'(w(x)) only because it would be analogous to the case with finitely many possible value of x, but I dont know how to prove that.

2. Originally Posted by southprkfan1
Let x be a random variable with density function f and [a,b] be the possible set of values of x.

Let w(x) be a value associated with each value of x. I.e. w() is not known, and suppose that U(w(x)) is a $C^{\infty}$ function.

Let $H(w(x)) = \int_{a}^b f(x)U(w(x))dx$

Find the integral of H with respect to w over all x.

I don't know where to begin because I'm not even sure I understand the question.

I suspect the answer is f(x)U'(w(x)) only because it would be analogous to the case with finitely many possible value of x, but I dont know how to prove that.
Are you to find the integral of H or the derivative?

The derivative is, using the fundamental theorem of calculus and the chain rule, f(x)U'(w(x)) as you say. The integral would be much messier.

3. Originally Posted by HallsofIvy
Are you to find the integral of H or the derivative?

The derivative is, using the fundamental theorem of calculus and the chain rule, f(x)U'(w(x)) as you say. The integral would be much messier.
Oops, Yes I meant the derivative of H. Could you explain how you get the answer in a little more detail.

4. Originally Posted by southprkfan1
Oops, Yes I meant the derivative of H. Could you explain how you get the answer in a little more detail.
Specificially, wouldn't the FTOC imply the answer should be: f(x)U(w(x))*U'(W(x))?