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Say I am sampling $x \in \mathbb{R}^2$ from a nice (uniform!) distribution which is supported on the unit disk supported on the origin. Given this is there a way to sample another random vector $y \in \mathbb{R}^2$ such that at least $\mathbb{E} [x_i y_i] = p$ for $i \in \{1,2\}$ for some given constant $p$ and maybe also ensure that the tuples $(x_i,y_i)$ for each $i$ are $i.i.d$?

  • Feel free to change the distribution of $x$ to something else if that helps you give me such an example!

  • I am happy to see even examples where just the first condition $\mathbb{E} [x_i y_i] = p$ is satisfied.

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    Seems a bit vague. Do you have a value of $p$ in mind. What is the motivation for $y?$ – 2017-01-17
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    $p$ is some given real constant. I am assuming that the construction of this $y$ will just treat it as a parameter and nothing depends on its precise value. – 2017-01-18

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For the $x_i$ you gave, $\mathbb{E}[x_i] = 0$, and so $x$ and $y$ being i.i.d. implies that $\mathbb{E}[x_iy_i] = 0$.

If you drop the i.i.d. requirement, then consider $y_i = p/x_i$, which gives $\mathbb{E}[x_i y_i] = p$ (this is a bad idea). A better choice would be $y_i = \frac{p}{\mathbb{E}[x_i^2]} x_i$, which gives $\mathbb{E}[x_iy_i] = p\mathbb{E}[x_i^2]/\mathbb{E}[x_i^2] = p$.

If you want to keep the i.i.d. requirement, then consider $x$ and $y$ sampled uniformly from the disk centered on $(\sqrt{p},\sqrt{p})$. By the iid assumption, $\mathbb{E}[x_iy_i] = \mathbb{E}[x_i]\mathbb{E}[y_i] = \sqrt{p}^2 = p$.

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    Thanks! Something is not clear to me. Isn't defining $y_i = p/x_i$ problematic because $x_i = 0$ is a possibility although a measure $0$ possibility? Further if $x$ (or $y$) is sampled uniformly from a disk then don't their coordinates become dependent or at least correlated and hence the coordinates aren't i.i.d anymore? – 2017-01-19
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    You're right about $y_i = p/x_i$ not being quite right - I don't think it would be too hard to find another correlated distribution. A better example would be $y_i = \frac{p}{\mathbb{E}[x_i^2]} x_i$. If you sample $x$ and $y$ _independently_ from the $\sqrt{p}$-centered disk, they are independent and identically distributed, with $\mathbb{E}[x_i] = \mathbb{E}[y_i] = \sqrt{p}$. – 2017-01-19
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    In the second case why are the coordinates of the vector $x$ or $y$ mutually independent random variables? – 2017-01-20
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    One equivalent definition of independence is that $p(x_i \mid y_i) = p(x_i)$. Now given that you know $y_i$, does this change the distribution of $x_i$? There is no information about the value of $x_i$ encoded in $y_i$, so it shouldn't. – 2017-01-20
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    I think you are confusing my question. Take two coordinates of $y$ say $y_i$ and $y_j$ for $i \neq j$ (and similarly $x_i$ and $x_j$. In your centered disk example it doesn't seem that either of the pairs $(x_i,x_j)$ or $(y_i,y_j)$ is mutually independent. – 2017-01-20
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    I see what you are saying now. If $x \in \mathbb{R}^2$ is a vector with two coordinates sampled from a disk, this intrinsically means that the coordinates are correlated, since $x_1 = r$ implies $x_2 = 0$. If you sample from a square centered at $\sqrt{p},\sqrt{p}$, then $x_1,x_2,y_1,y_2$ are all independent, identically distributed random variables, and so $(x_1,y_1)$ is mutually independent from $(x_2,y_2)$. Also, using the same reasoning above we have $\mathbb{E}[x_iy_i] = p$. If this is the answer you were looking for I'll update my post. – 2017-01-20