/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 76 Let \(X\) and \(Y\) be independe... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 76

Let \(X\) and \(Y\) be independent random variables with means \(\mu_{x}\) and \(\mu_{y}\) and variances \(\sigma_{x}^{2}\) and \(\sigma_{y}^{2}\). Show that $$ \operatorname{Var}(X Y)=\sigma_{x}^{2} \sigma_{y}^{2}+\mu_{y}^{2} \sigma_{x}^{2}+\mu_{x}^{2} \sigma_{y}^{2} $$

Short Answer

Expert verified
To show that \( Var(XY) = \sigma_{x}^{2} \sigma_{y}^{2} + \mu_{y}^{2} \sigma_{x}^{2} + \mu_{x}^{2} \sigma_{y}^{2} \), we first find \( E[XY] \) using the independence of X and Y: \( E[XY] = \mu_{x} \mu_{y} \). Then, we find the second moment \( E[(XY)^{2}] = (\sigma_{x}^{2}+\mu_{x}^{2})(\sigma_{y}^{2}+\mu_{y}^{2}) \). Using the variance formula, \( Var(XY) = E[(XY)^{2}] - (E[XY])^{2} \), and simplifying, we arrive at the desired result.

Step by step solution

01

Find the expectation \( E[XY] \)

Since X and Y are independent, we can use the property \( E[XY]=E[X]E[Y] \) for independent random variables: \( E[XY] = E[X]E[Y] = \mu_{x} \mu_{y} \)
02

Find the second moment \( E[(XY)^{2}] \)

Again, using the independence of X and Y, we can find the second moment of XY: \( E[(XY)^{2}] = E[X^{2} Y^{2}] = E[X^{2}]E[Y^{2}] \) Now, let's find \( E[X^{2}] \) and \( E[Y^{2}] \) separately: \( E[X^{2}] = Var(X) + (E[X])^{2} = \sigma_{x}^{2} + \mu_{x}^{2} \) \( E[Y^{2}] = Var(Y) + (E[Y])^{2} = \sigma_{y}^{2} + \mu_{y}^{2} \) Now, substitute these back into the formula for \( E[(XY)^{2}] \): \( E[(XY)^{2}] = (\sigma_{x}^{2}+\mu_{x}^{2})(\sigma_{y}^{2}+\mu_{y}^{2}) \)
03

Calculate the variance of XY using the variance formula

The variance formula is: \( Var(XY) = E[(XY)^{2}] - (E[XY])^{2} \) Plug in the expressions we found for \( E[XY] \) and \( E[(XY)^{2}] \): \( Var(XY) = (\sigma_{x}^{2}+\mu_{x}^{2})(\sigma_{y}^{2}+\mu_{y}^{2}) - (\mu_{x} \mu_{y})^{2} \)
04

Simplify the expression to get the desired result

Now, expand the expression and simplify: \( Var(XY) = \sigma_{x}^{2} \sigma_{y}^{2} + \mu_{y}^{2} \sigma_{x}^{2} + \mu_{x}^{2} \sigma_{y}^{2} + \mu_{x}^{2} \mu_{y}^{2} - \mu_{x}^{2} \mu_{y}^{2} \) The terms with \( \mu_{x}^{2} \mu_{y}^{2} \) cancel each other out: \( Var(XY) = \sigma_{x}^{2} \sigma_{y}^{2} + \mu_{y}^{2} \sigma_{x}^{2} + \mu_{x}^{2} \sigma_{y}^{2} \) Thus, we have shown the desired result.

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Most popular questions from this chapter

A coin having probability \(p\) of coming up heads is successively flipped until the \(r\) th head appears. Argue that \(X\), the number of flips required, will be \(n, n \geq r\), with probability $$ P[X=n\\}=\left(\begin{array}{c} n-1 \\ r-1 \end{array}\right) p^{T}(1-p)^{n-r}, \quad n \geq r $$ This is known as the negative binomial distribution. Hint: How many successes must there be in the first \(n-1\) trials?

Suppose that \(X\) and \(Y\) are independent binomial random variables with parameters \((n, p)\) and \((m, p) .\) Argue probabilistically (no computations necessary) that \(X+Y\) is binomial with parameters \((n+m, p)\).

Prove that \(E\left[X^{2}\right] \geq(E[X])^{2}\). When do we have equality?

Suppose that we want to generate a random variable \(X\) that is equally likely to be either 0 or 1 , and that all we have at our disposal is a biased coin that, when flipped, lands on heads with some (unknown) probability \(p\). Consider the following procedure: 1\. Flip the coin, and let \(0_{1}\), either heads or tails, be the result. 2\. Flip the coin again, and let \(0_{2}\) be the result. 3\. If \(0_{1}\) and \(0_{2}\) are the same, return to step 1 . 4\. If \(0_{2}\) is heads, set \(X=0\), otherwise set \(X=1\). (a) Show that the random variable \(X\) generated by this procedure is equally likely to be either 0 or 1 . (b) Could we use a simpler procedure that continues to flip the coin until the last two flips are different, and then sets \(X=0\) if the final flip is a head, and sets \(X=1\) if it is a tail?

Calculate the moment generating function of the uniform distribution on \((0,1)\). Obtain \(E[X]\) and \(\operatorname{Var}[X]\) by differentiating.
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