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I am trying to calculate the following conditional expectation.

Let Z be a uniformly distributed random variable on the closed interval $[0, 2]$.
Define $X = \cos(\pi Z)$ and $Y = 2\sin(\pi Z)$.
Calculate $\mathbb{E}[Y|X] = \mathbb{E}[Y|\sigma(X)]$.

I have tried multiple approaches, but don't know how to proceed since I am only familiar with the standard techniques for calculating the conditional expectation, i.e. when Y is independent of X, Y is measurable with respect to $\sigma(X)$ or if a joint density $f(x,y)$ exists. The first two parts don't apply here and since X and Y aren't independent I don't know how to calculate the joint density.

Any help is greatly appreciated. Thank you very much!

S. Catterall
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Matija
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3 Answers3

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There are at least three approaches to this problem.

  1. Intuition. There is no joint density of $X$ and $Y$ with respect to the usual Lebesgue measure on $\mathbb{R}^2$. However, it should be clear that $(X,Y/2)$ is uniformly distributed on the unit circle in $\mathbb{R}^2$. The unit circle is symmetric in both the $x$ and $y$ axes of the plane, so it follows that $\mathbb{E}[Y|X]=\mathbb{E}[X|Y]=0$.

  2. Kolmogorov conditional expectation. You can try the approach described in this answer, which is as follows. We know that $\mathbb{E}[Y|X]=g(X)$ for some Borel measurable function $g$. Now use the defining property of the conditional expectation to pin down what $g$ actually is. I haven't worked out all the details but I think this approach should work.

  3. Regular conditional probabilities. This is the approach hinted at by @AntonioSpeltzu but without much detail. It is more intuitive than the second approach, but requires the additional machinery of regular conditional probabilities, as described here (general theory) and here (how to use the theory in practice). They allow us to give meaning to statements such as $$P\left(Y/2=\frac{1}{2} | X=\frac{\sqrt{3}}{2}\right)=\frac{1}{2}$$ This approach will get you the right answer, but is not strictly necessary because the question only asks about conditional expectations rather than condtional probabilities.

S. Catterall
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This is a simplified case from:
https://stats.stackexchange.com/questions/603110/how-to-calculate-px-in-a-sinx-y

Here: $P[Z\in A|X=x]=\frac{1}{2}(I_A(arccos(x)/\pi)+I_A((2\pi-arccos(x))/\pi))$
Therefore: $E[X|Y]=E[Y|X]=0$

Speltzu
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I have an upcoming exam, so I wanted to tackle this question as an exercise. We use the second approach suggested by the accepted answer.

Let $X=\cos(\pi Z)$ and $Y=\sin(\pi Z)$ where $Z \sim \text{Uniform}([0,2])$. Let $h(X)$ be a version of the conditional expectation $\mathbb{E}[Y|X]$. Let $u : \mathbb{R} \rightarrow \mathbb{R}$ bounded and measurable (or non-negative and measurable, it does not matter for this approach). It must hold that

$$ \mathbb{E}[Y u(X)] = \mathbb{E}[h(X) u(X)]$$

With this in mind, we calculate:

$$ \begin{align*} \mathbb{E}[Yu(X)] &= \frac{1} {2\pi}\int_0^{2\pi}\sin(x)u(cos(x)) \; dx \\\\ &=\frac{1}{2\pi}\left( \int_0^{\pi}\sin(x)u(cos(x)) \;dx + \int_{\pi}^{2\pi}\sin(x)u(cos(x)) \;dx \right) \\\\ &=\frac{1}{2\pi}\left( \int_0^{\pi}\sin(x)u(cos(x)) \;dx + \int_0^\pi\sin(2\pi - x)u(cos(2\pi - x)) \;dx \right) \\\\ &=\frac{1}{2\pi}\left( \int_0^{\pi}\sin(x)u(cos(x)) \;dx + \int_0^\pi-\sin(x)u(cos(x)) \;dx \right) \\\\ &=0 \end{align*} $$

Since this holds for all $u$, it follows that $h(X)=0$ a.e.

Ata Keskin
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