Timeline for How to deal with the vector norm item as a denominator in this expectation?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 30, 2013 at 15:15 | comment | added | Pait | Is it so clear that $R$ and $Y$ are independent? | |
| Dec 30, 2013 at 14:00 | comment | added | Stanley Yao Xiao | Your question is not clear. What do you expect to do with the expression $E(1/x)$? Do you want it as a function of $E(x)$? | |
| Dec 30, 2013 at 13:18 | comment | added | Shan | Hello I just want to know how you applied expectation in denominator? As I want to apply expectation in one of the well-known formula to find out its unbiasedness. the expression is E(1/x)=? | |
| Dec 14, 2011 at 2:25 | comment | added | ppyang | fedja, it's so kind of you to help me so much. Your idea is fantastic and I think this also provides a simple approach to solve the problem I had proposed before at mathoverflow.net/questions/81419/… Thank you very much! | |
| Dec 14, 2011 at 0:34 | comment | added | fedja | $X=RY$ where $Y$ is equidistributed over the unit sphere and $R=\|X\|$. Note that $R$ and $Y$ are independent. Now, let $W(X)$ be your numerator. Then $W(X)=R^4W(Y)$ and, by independence, $EW(X)=EW(Y)ER^4$. Also $\frac{W(X)}{R^2}=R^2W(Y)$ whence the expectation you want is $EW(Y)E(R^2)$. Now just juxtapose these two formulae and recall that $ER^k=E\|X\|^k$. And yes, if you put $\|X\|^4$, you, indeed, get pure $EW(Y)$, so you get what you wrote :) | |
| Dec 13, 2011 at 16:01 | comment | added | ppyang | fedja, could you please explain your suggestion in more detail? Did you say that the degree is 2 just because 4(degree of numerator)-2(degree of denominator)=2? Think about the problem if I change the $\|X\|^2$ with $\|x\|^4$, and its expectation is just $1/E\|X\|^4$ times the expectation of the numerator? Thank you! | |
| Dec 13, 2011 at 15:40 | comment | added | ppyang | fedja, I have not made sense of your explanation, but I think you suggested a good way to think about this problem. I will think about it with your help. Thank you very much! | |
| Dec 13, 2011 at 12:57 | comment | added | fedja | You have the spherical symmetry and the radial homogeneity of degree $2$ as opposed to degree $4$ without the denominator. So the expectation of the fraction is just $(E\|X\|^2)/E(\|X\|^4)$ times the expectation of the numerator. | |
| Dec 13, 2011 at 7:24 | history | asked | ppyang | CC BY-SA 3.0 |