The value of $correlation(X, Y)$ is __________ . (Answer in integer)
We need to find the correlation between random variables $X$ and $Y$, denoted as $Corr(X, Y)$.
First, let's find the expected value of $X$ and $Y$ given $X=x$.
Using the law of total expectation, $E[Y] = E[E[Y|X]]$:
$E[Y] = E[X^2]$
For $X \sim Uniform(a, b)$, $E[X^2] = \frac{a^2 + ab + b^2}{3}$.
Here, $a = -1$ and $b = 1$.
$E[X^2] = \frac{(-1)^2 + (-1)(1) + (1)^2}{3} = \frac{1 - 1 + 1}{3} = \frac{1}{3}$.
Therefore, $E[Y] = \frac{1}{3}$.
The formula for covariance is $Cov(X, Y) = E[XY] - E[X]E[Y]$.
We need $E[XY]$. Using the law of total expectation again:
$E[XY] = E[E[XY|X]] = E[X \cdot E[Y|X]]$
Substitute $E[Y|X] = X^2$:
$E[XY] = E[X \cdot X^2] = E[X^3]$.
For $X \sim Uniform(-1, 1)$, $E[X^k] = \frac{1}{1 - (-1)} \int_{-1}^{1} x^k dx$.
Calculate $E[X^3]$:
$E[X^3] = \frac{1}{2} \int_{-1}^{1} x^3 dx = \frac{1}{2} \left[ \frac{x^4}{4} \right]_{-1}^{1} = \frac{1}{2} \left( \frac{1^4}{4} - \frac{(-1)^4}{4} \right) = \frac{1}{2} \left( \frac{1}{4} - \frac{1}{4} \right) = 0$.
So, $E[XY] = 0$.
Now, calculate the covariance:
$Cov(X, Y) = E[XY] - E[X]E[Y] = 0 - (0) \cdot (\frac{1}{3}) = 0$.
The correlation is calculated as $Corr(X, Y) = \frac{Cov(X, Y)}{\sqrt{Var(X) Var(Y)}}$.
Since $Cov(X, Y) = 0$, and the variances $Var(X)$ and $Var(Y)$ are finite and non-zero (specifically $Var(X) = \frac{(1-(-1))^2}{12} = \frac{4}{12} = \frac{1}{3}$), the correlation is:
$Corr(X, Y) = \frac{0}{\sqrt{Var(X) Var(Y)}} = 0$.
The value of $correlation(X, Y)$ is 0.
The value of simple correlation coefficient lies in the interval:
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