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I'm reading an algorithm proof in which there is a passage I can't understand. Substantially we have 2 statistically independent random arrays $X$ and $Z$, and a third random array $Y$. It's known that $Y=XZ$ and $E[Z] = 1$. What the author provides are textually the following passages: \begin{align} VAR[Y] &= E[(Y -E[Y])^2]\\ &= E[(X(Z-1) + (X-E[X]))^2]\\ &= (VAR[X]+E[X^2])VAR[Z]+VAR[X] \end{align} where $E[X]$ and $VAR[X]$ are the expected value (i.e. mean) and the variance of $X$, respectively.

Trying to understand that proof, I'm getting a different result, as it is: \begin{align} VAR[Y] &= E[(Y -E[Y])^2]\\ &= E[(XZ - E[XZ])^2]\\ &= E[(XZ -X +X -E[X])^2]\\ &= E[(X(Z-1) + (X-E[X]))^2]\\ &= E[X^2(Z-1)^2 + (X - E[X])^2 + 2X(Z-1)(X - E[X])]\\ &= E[X^2]E[(Z-1)^2] + VAR[X] + 2E[X(Z-1)(X - E[X])]\quad \text{(then, as $E[Z] = 1$)}\\ &= E[X^2]VAR[Z] + VAR[X] + 2E[X^2Z - ZXE[X] - X^2 + XE[X]]\\ &= E[X^2]VAR[Z] + VAR[X] + 2E[X^2Z] - 2E[ZXE[X]] - 2E[X^2] + 2E[XE[X]]]\\ &= E[X^2]VAR[Z] + VAR[X] + 2E[X^2]E[Z] - 2E[XE[X]]E[Z] - 2E[X^2] + 2E[XE[X]]]\\ &= E[X^2]VAR[Z] + VAR[X] + 2E[X^2] - 2E[XE[X]] - 2E[X^2] + 2E[XE[X]]]\\ &= E[X^2]VAR[Z] + VAR[X] \neq (VAR[X]+E[X^2])VAR[Z]+VAR[X] \end{align}

Can you help me to understand where am I wrong? Thank you very much in advance for your attention.

EDIT: This is the original author proof algorithm proof

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There is a misprint in your reproduction of the author's final formula, where $E(X^2)$ should read $E(X)^2$. Then the author's formula and yours become equivalent (and both correct). –  Did Aug 25 '12 at 8:47
    
Ok, I added the original author proof passage (5.3). Can you give me a more detailed explaination of the formulae equivalence? thank you in advance. –  Alfatau Aug 25 '12 at 9:03
    
It appears that the misprint is yours and that the author wrote a correct formula: you wrote $E(X^2)$ for $\bar x^2=E(X)^2$. // The equivalence between the result of your computations and what is (really) written in the book is clear since $\mathrm{Var}(X)+E(X)^2=E(X^2)$. –  Did Aug 25 '12 at 9:11
    
I'm sorry but I still don't understand. Why do you say mine is a misprint? if you follow the passages it should really be $E[X^2]$ and not $E[X]^2$, or have I done something wrong in my calculations? thank you again for your attention. –  Alfatau Aug 25 '12 at 9:16
1  
@Alfatau: I don't see any problem with the author's notation. It seems to me that you led yourself to a wrong interpretation of the formula. –  joriki Aug 25 '12 at 9:26

1 Answer 1

up vote 3 down vote accepted

I don't see in what sense $X$, $Y$ and $Z$ are arrays, especially given $E[Z]=1$; I'll assume that they're just random variables. (Also where you write "pass" it seems you mean "passage".)

Your calculation is correct. You can check this using a simple example, for instance $X$ and $Z$ both uniformly drawn from $\{0,2\}$.

You could considerably shorten your calculation by noting that the expected value of any term of the form $f(X)(Z-1)$ vanishes; that gets rid of the mixed term that the last few lines focus on.

share|improve this answer
    
Yes. And see my comment. –  Did Aug 25 '12 at 8:48
    
ok, thanks. $X$,$Y$,$Z$ are random arrays and not random variables because they represent sampled sequences of data and each element of $Z$ has an exponential distribution with the same mean and standard deviation. In the proof it has been initially supposed $E[Z] = 1$. however, I still don't understand why my result is different from the author's one. Any idea about it? –  Alfatau Aug 25 '12 at 8:51
    
@Alfatau: What you write about arrays and variables appears to indicate a confusion on your part. The mean of the values in a random array can be $1$, but the expectation of a random array is an array. You can have an array of samples drawn from a random variable, but that doesn't turn the random variable into an array, and it doesn't change the fact that $Z$ and $E[Z]$ necessarily have the same dimensions, so if $E[Z]$ is a number, then $Z$ is a random variable. –  joriki Aug 25 '12 at 9:25

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