Maximizing over a continuum of possibilities

Feb 2010
147
32
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
Apr 2005
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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.
 
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Feb 2010
147
32
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.