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Q.26

Expert-verified

A First Course in Probability

Found in: Page 361

Book edition 9th

Author(s) Sheldon M. Ross

Pages 432 pages

ISBN 9780321794772

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Short Answer

Prove that $E [g (X) Y ∣ X] = g (X) E [Y ∣ X]$ .

We prove that, $E [g (X) Y ∣ X] = g (X) E [Y ∣ X]$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given information

Given in the question that, we have to prove that $E [g (X) Y ∣ X] = g (X) E [Y ∣ X]$

Step 2: Explanation

Because of the clarity, suppose that $X$ and $Y$ are continuous random variables with its density functions. Take any $x \in$ $supp f_{X}$ . We have that

$E (g (X) Y ∣ X = x) = \iint_{ℝ^{2}} g (x) y f_{(X, Y) ∣ X = x} (x, y) d x d y$

But, we have that

$f_{(X, Y) ∣ X = x} (x, y) d x d y = f_{Y ∣ X} (y ∣ x) d y$

So the integral becomes

$\int_{ℝ} g (x) y f_{Y ∣ X} (y ∣ x) d y$

Which is equal to

$g (X) E (Y ∣ X = x)$

Since the relation hold for every $x \in supp f$ , we end up with

$E (g (X) Y ∣ X) = g (X) E (Y ∣ X)$

Step 3: Final answer

We prove that, $E [g (X) Y ∣ X] = g (X) E [Y ∣ X]$

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