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I want to know what probability distribution has the linearity property of the conditional expectation.

To be specific, suppose that we have three random variables named $v_1,\;v_2,\;v_3$.

Then, if $[v_1,\;v_2,\;v_3]$ follows a joint normal distribution, we can show that $\mathbb{E}[v_1|v_2,\;v_3]$ is linear in $v_2$ and $v_3$.

That is, $\mathbb{E}[v_1|v_2,\;v_3]=\rho'[v_2,\;v_3]'$, where $\rho$ is a $2\times1$ constant vector consists of covariance of $[v_1,\;v_2,\;v_3]$.

Here, according to a textbook, there are diverse probability distributions that have the same linearity of the conditional expectation.

But, I am not sure what distributions also have that property.

Could you give me some examples (with proof if possible).

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    $\begingroup$ There is a huge number of such distributions one can construct. See stats.stackexchange.com/questions/257779 for an example. $\endgroup$
    – whuber
    Jul 10 at 16:22
  • $\begingroup$ @whuber Thank you for your comment every time. Your answers make me inspired and are always informative :) $\endgroup$
    – M.C. Park
    Jul 11 at 4:03

2 Answers 2

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Multivariate Elliptical distributions deals with linearity in conditional expectation. You can think about this family as a generalization of Normal distribution, t-Student is another notable example of this family.

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Suppose that jointly continuous random variables $X$ and $Y$ have joint pdf with constant value $2$ on the triangle with vertices $(0,0), (1,1), (0,1)$ in the plane. Note that this is the joint pdf of $X = \min(U,V)$ and $Y=\max(U,V))$ where $U$ and $V$ are i.i.d, $\mathcal U(0,1)$ random variables. Then, $E[X\mid Y] = \frac 12 Y$ and $E[Y\mid X] = \frac 12 + \frac 12 X$ both are linear functions of $Y$ and $X$ respectively.

As to whether the joint distribution of the minimum and maximum of two independent $\mathcal U(0,1)$ random variables is "a typical well-known distribution" or not is something that I leave for the cognoscenti on this group to decide. I am just an engineer and a mere dabbler in statistics, not a statistician

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  • $\begingroup$ So, your point is that the linearity could hold even though the joint distribution is not a typically well-known one. Right? $\endgroup$
    – M.C. Park
    Jul 9 at 15:04
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    $\begingroup$ @M.C.Park You asked "Could you give me some examples (with proof if possible)?" and so I gave you an example. I didn't prove that $E[X\mid Y]$ and $E[Y\mid X]$ are linear functions as you want, but then, neither did the answer that you have accepted. $\endgroup$ Jul 10 at 11:42
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    $\begingroup$ I doubt I count among any cognoscenti, but for what it's worth I'd call that distribution "well-known" (and the claimed property is so starkly obvious in this case that it is a perfect example to give). $\endgroup$
    – Glen_b
    Jul 10 at 22:49
  • $\begingroup$ @DilipSarwate I think, I made a misleading point in my question by using the term "example". That is, the word "example" is ambiguous in the question. What I mean by "example" was "a family of distributions" such as normal, logistic, and so on. That is the reason why I accepted the first answer. I am sorry for my use of the vague term. $\endgroup$
    – M.C. Park
    Jul 11 at 3:59
  • $\begingroup$ @Glen_b Your false modesty cuts no ice with me or with any other readers of this forum for that matter!. Your opinions and kind words are most valued. $\endgroup$ Jul 11 at 4:00

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