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Chapter 7: Problem 35

(a) Prove that $$ E[X]=E[X \mid X0$ $$ P\\{X \geq a\\} \leq \frac{E[X]}{a} $$

Short Answer

Expert verified
To prove part (a), let's define a new random variable \(Y\) as: \[ Y = \begin{cases} a & \text{if } X \geq a \\ X & \text{if } X < a \end{cases} \] Now, we can express \(E[X]\) as the sum of the two conditional expectations (\(E[X \mid X 0\): \[P\{X \geq a\} \leq \frac{E[X]}{a}\] By following through the inequalities and rearranging them, we can prove Markov's Inequality.

Step by step solution

01

Define a random variable

First, let's define a new random variable \(Y\) as: \[ Y = \begin{cases} a & \text{if } X \geq a \\ X & \text{if } X < a \end{cases} \] Now, we can write the expectation of \(X\) as: \[E[X] = E[X \cdot I\{X<a\} + Y \cdot I\{Y \geq a\}]\], where \(I\{...\}\) denotes the indicator function.
02

Calculate \(E[X \mid X

To calculate \(E[X \mid X<a]\) and \(E[X \mid X \geq a]\), we can use the definition of conditional expectation: \[E[X \mid X<a] = \frac{E[X \cdot I\{X<a\}]}{P\{X<a\}}\] and \[E[X \mid X \geq a] = \frac{E[Y \cdot I\{Y \geq a\}]}{P\{X \geq a\}}\]
03

Prove Part (a)

Now, we can express \(E[X]\) as the sum of the two conditional expectations multiplied by their respective probabilities, according to our definition of \(Y\): \[E[X] = E[X \cdot I\{X<a\}] + E[Y \cdot I\{Y \geq a\}]\] \[E[X] = E[X \mid X<a] \cdot P\{X<a\} + E[X \mid X \geq a] \cdot P\{X \geq a\}\] This proves part (a).
04

Prove Markov's Inequality

Now, let's use the result from part (a) to prove Markov's inequality, which states that if \(P\{X \geq 0\} = 1\), then for \(a > 0\): \[P\{X \geq a\} \leq \frac{E[X]}{a}\] From part (a), we have: \[E[X]=E[X \mid X<a] P\{X<a\}+E[X \mid X \geq a] P\{X \geq a\}\] Since \(X \geq 0\), we have \(E[X \mid X<a] \geq 0\). Also, when \(X \geq a\), we have \(E[X \mid X \geq a] \geq a\). Hence, \[E[X] \geq E[X \mid X<a] P\{X<a\} + a \cdot P\{X \geq a\}\] Now, we rearrange the inequality: \[P\{X \geq a\} \leq \frac{E[X] - E[X \mid X<a] P\{X<a\}}{a}\] Since the numerator is non-negative, the inequality holds: \[P\{X \geq a\} \leq \frac{E[X]}{a}\] This proves the Markov's inequality as stated in part (b).

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Most popular questions from this chapter

Um 1 contains 5 white and 6 black balls, while um 2 contains 8 white and 10 black balls. Two balls are randomly selected from um 1 and are then put. in urn 2 . If 3 balls are then randomly selected from urn 2 , compute the expected number of white balls in the trio. HNT: Let \(X_{i}=1\) if the \(i\) th white ball initially in urn 1 is one of the three selected, and let \(X_{i}=0\) otherwise. Similarly, let \(Y_{i}=1\) if the ith white ball from urn 2 is one of the three selected, and let \(Y_{i}=0\) otherwise. The number of white balls in the trio can now be written as \(\sum_{1}^{5} X_{i}+\sum_{1}^{8} Y_{i}\).

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