0
$\begingroup$

If $X,Y$ and $Z$ have identical distributions and are independent, and can assume $0$ or $1$, what is $\text{Var}(XYZ)$? So this is either $0$ or $1$. It's $1$ with probability $1/8$ and $0$ with probability $7/8$. This means that $\text{Var}(XYZ) = 7/16$? Or is this not right? How does this use the definition of variance? We could note that $E(X) = E(X^2)$, etc....?

$\endgroup$
1
  • 2
    $\begingroup$ I see from your profile that you have asked 27 questions on this site, yet your total number of upvotes is 1. You also haven't accepted answers to any of the last 10 questions you've asked. Just so you know, on this web site it's considered polite - as it shows respect and appreciation for the people who take the time to answer your questions - both to upvote the answers that you find helpful and to formally accept the answer to each question that you consider to be the best. $\endgroup$ Nov 16, 2010 at 3:50

2 Answers 2

3
$\begingroup$

$Var(X) = E[X^2] - E[X]^2$. Assuming each of the variables take 0 or 1 equally likely, as you noted, $E[XYZ] = E[X^2Y^2Z^2] = 1/8$. So $Var(X) = 1/8 - 1/64 = 7/64$.

$\endgroup$
2
$\begingroup$

You imply that $X, Y$, and $Z$ have equal chance of being 0 or 1. In that case $E((XYZ)^2)=\frac{1}{8}$ and $(E(XYZ))^2=\frac{1}{64}$. So the variance would be $\frac{7}{64}$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .