Show that if X has pmf p(x), thenE[h(X)E[Y |X]] = ∑xh(x)E[Y |X = x]p(x)
Question
Show that if X has pmf p(x), thenE[h(X)E[Y |X]] = ∑xh(x)E[Y |X = x]p(x)
Solution
The statement you're asking to prove is a property of the expected value in the context of probability theory and statistics. Here's a step-by-step proof:
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Start with the definition of conditional expectation: E[Y|X] is a random variable whose value is E[Y|X = x] with probability p(x).
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Therefore, h(X)E[Y|X] is also a random variable whose value is h(x)E[Y|X = x] with probability p(x).
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The expected value of a random variable is the sum of the possible values it can take, each multiplied by their respective probability.
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Therefore, E[h(X)E[Y|X]] = ∑xh(x)E[Y|X = x]p(x).
This completes the proof.
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