Timeline for Separating $X$ from $Y$ in $E[(X^T Y))^p]$ for $p = 3$ and $4$?
Current License: CC BY-SA 4.0
13 events
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| Dec 5, 2020 at 13:43 | comment | added | πr8 | Again - in principle, such an expression exists, but in practice, it is unlikely to be easy to work with. Inevitably, you will end up either having to do some combinatorics or some advanced linear algebra (e.g. using tensors). | |
| Dec 5, 2020 at 12:50 | comment | added | Bertus101 | Ah ok I see now. What about the case when they are not iid as that is actually what I am interested in? In your answer to my previous question the fact that you proved $E[(X^T(Y-Z))^2] = Tr[XX^T]E[(Y-Z)(Y-Z)^T]$ was very useful to me as when $Z$ is the mean of $Y$, I was able to use $E[(Y-Z)(Y-Z)^T] = \text{Cov}[Y,Y]$ and account for the covariances between the elements of $Y$. So do you know if there is a nice formula for the non-iid $P=3$ case? | |
| Dec 5, 2020 at 11:55 | comment | added | πr8 | I have made the assumption that the coordinates of $X$ and $Y$ are each iid, so $\mathbf{E} \left[ X^3 \right] = \mathbf{E} \left[ X_1^3 \right]$, etc. | |
| Dec 5, 2020 at 10:22 | vote | accept | Bertus101 | ||
| Dec 5, 2020 at 10:22 | comment | added | Bertus101 | How do we interpret powers of random vectors such as $E[X^3]$. Does this notation mean $E[X^3] = E[X^T X X^T]$ so that $E[X^3]$ would then be a random vector and $E[X^3]\cdot E[Y^3]$ would be a dot product of two random vectors? Similarly with $E[X]^3$, does this mean $E[X]^T E[X] E[X]^T$? | |
| Dec 4, 2020 at 23:54 | comment | added | Ben | @Bertus101: Finite summations of finite products is already "closed form" (though not particulary simple looking in this case). | |
| Dec 4, 2020 at 20:35 | history | edited | πr8 | CC BY-SA 4.0 |
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| Dec 4, 2020 at 19:40 | comment | added | Bertus101 | The pattern for general $P$ is very interesting but my main concern is the $P=3$ and $4$ cases. I'm wondering are there 'nice' expressions for these cases like the $P=1$ and $2$ cases. | |
| Dec 4, 2020 at 17:44 | history | edited | πr8 | CC BY-SA 4.0 |
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| Dec 4, 2020 at 16:53 | comment | added | whuber♦ | This general result simplifies by means of multinomial coefficients. That reduces the expression to sums over integer partitions of $P.$ This rapidly gets more and more complicated as $P$ increases. | |
| Dec 4, 2020 at 16:35 | history | edited | StatsStudent | CC BY-SA 4.0 |
fixed typos.
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| Dec 4, 2020 at 16:01 | comment | added | Bertus101 | Do you know if there are nice closed form expressions for the $P=3$ and $4$ cases? | |
| Dec 4, 2020 at 15:52 | history | answered | πr8 | CC BY-SA 4.0 |