# Maximizing over a continuum of possibilities

#### 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 $$\displaystyle C^{\infty}$$ function.

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

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.

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#### HallsofIvy

MHF Helper
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 $$\displaystyle C^{\infty}$$ function.

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

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.

• southprkfan1

#### southprkfan1

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.

#### 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))?