Question

Let \(X\) be a random variable that follows \(Uniform(-1,1)\) distribution. The conditional distribution of the random variable \(Y\) given \(X = x\) is the \(Uniform(x^2-0.1, x^2+0.1)\) distribution.

The value of \(correlation(X,Y)\) is _______. (Answer in integer)

Hint:

Use the tower property: \(E[XY] = E[X \cdot E[Y|X]] = E[X^3]\), and note that \(X^3\) is an odd function of a symmetric variable, so its expectation is 0, making the covariance and hence the correlation 0.

Answer 1

Correct Answer: 0

Step 1: Write down what we know about X and the conditional distribution of Y.
\(X \sim Uniform(-1,1)\), so its density is symmetric about 0. This means \(E[X] = 0\), and also that the distribution of \(X\) is symmetric: \(X\) and \(-X\) have the same distribution.
Given \(X = x\), \(Y\) is drawn from \(Uniform(x^2-0.1, x^2+0.1)\), an interval of width 0.2 centered at \(x^2\).

Step 2: Find the conditional mean of Y.
For a uniform distribution, the mean is the midpoint of the interval. So \[ E[Y \mid X=x] = \frac{(x^2-0.1)+(x^2+0.1)}{2} = x^2 \]

Step 3: Recall the formula for correlation and what we need. \[ correlation(X,Y) = \frac{Cov(X,Y)}{\sigma_X \sigma_Y} \] We only need the value of \(Cov(X,Y)\), because if it turns out to be 0, the correlation is 0 as long as \(\sigma_X\) and \(\sigma_Y\) are finite and nonzero, which they are here. \[ Cov(X,Y) = E[XY] - E[X]E[Y] \]

Step 4: Compute E[XY] using the tower property (law of total expectation). \[ E[XY] = E\big[E[XY \mid X]\big] = E\big[X \cdot E[Y\mid X]\big] = E[X \cdot X^2] = E[X^3] \] Now, \(X \sim Uniform(-1,1)\) is symmetric about 0, and \(X^3\) is an odd function. For a distribution symmetric about 0, the expectation of any odd function of \(X\) is 0, because the positive and negative contributions cancel exactly. Explicitly: \[ E[X^3] = \int_{-1}^{1} x^3 \cdot \frac{1}{2}\,dx = \frac{1}{2}\left[\frac{x^4}{4}\right]_{-1}^{1} = \frac{1}{2}\left(\frac{1}{4}-\frac{1}{4}\right) = 0 \] So \(E[XY] = 0\).

Step 5: Compute Cov(X,Y) and hence the correlation. We already have \(E[X] = 0\). So \[ Cov(X,Y) = E[XY] - E[X]E[Y] = 0 - (0)(E[Y]) = 0 \] Since the covariance is exactly 0, and both \(\sigma_X\) and \(\sigma_Y\) are finite positive numbers here, the correlation is \[ correlation(X,Y) = \frac{0}{\sigma_X \sigma_Y} = 0 \]

Final Answer:
The correlation between X and Y is 0. \[ \boxed{0} \]