Exercise 4.2 - Uncorrelated and Gaussian does not imply independent unless jointly Gaussian

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For question (a). The p.d.f. for $Y$ is:

p (Y \in [a, a + d a]) = 0.5 \cdot p (X \in [a, a + d a]) + 0.5 \cdot p (X \in [- a - d a], - a) = p (X \in [a, a + d a]),

since $X$ is symmetric. So $Y$ subject to a normal distribution $(0, 1)$ .

For question (b), we have:

\begin{align} cov (X, Y) = & 𝔼 (𝑋𝑌) - 𝔼 (X) - 𝔼 (Y) \\ = & 𝔼_{W} (𝔼 (𝑋𝑌 | W)) - 0 \\ = & 0.5 \cdot 𝔼 (X^{2}) + 0.5 \cdot 𝔼 (- X^{2}) = 0 . \end{align}

So they are uncorrelated.

To disprove dependence (in case of confusion), let:

a = Φ^{- 1} (\frac{1}{4}),

where $Φ$ is the c.d.f of $X$ , i.e.:

\int_{- \infty}^{a} 𝒩 (x | 0, 1) d x = \frac{1}{4} .

Let $R_{1} = (- \infty, a]$ , $R_{2} = (a, 0]$ . The space of experiment results for $X \times Y$ is $R^{2}$ . Let $R_{1} \times R_{2}$ be a Borel set in $R^{2}$ . Be $X$ and $Y$ independent, its probability measure should be $\frac{1}{16}$ . However, when $X \in R_{1}$ , it is impossible for $Y$ to take a value from $R_{2}$ . Hence the independency fails.

The rule of iterated expectation is but the Bayes rule:

\begin{aligned} 𝔼 [𝑋𝑌] & = \int_{X} \int_{Y} 𝑋𝑌 \cdot p (X, Y) d X d Y \\ = \int_{X} \int_{Y} 𝑋𝑌 \cdot (\int_{W} p (X, Y, W) d W) d X d Y \\ = \int_{W} (\int_{X} \int_{Y} 𝑋𝑌 \cdot p (X, Y | W)) p (W) d W . \end{aligned}

solour_lfq

2021-03-24 13:42

Exercise 4.2 - Uncorrelated and Gaussian does not imply independent unless jointly Gaussian

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