Question

Let X be a random variable that follows uniform (-1, 1) dist. The conditional dist. of the random variable Y given X = x is the Uniform (x$^2$ - 0.1, x$^2$ + 0.1) dist. The value of correlation (X, Y) is _______.

Hint:

A useful shortcut: If you can show that $E[Y|X]$ is an even function of X (like $X^2$ here), and the distribution of X is symmetric about 0 (like U(-1,1) or a standard Normal), then the covariance Cov(X,Y) will be 0. This is because $E[XY] = E[X \cdot E[Y|X]]$, and the expectation of an odd function ($X$ times an even function is odd) over a symmetric interval is zero.

Answer 1

Correct Answer: 0

Step 1: Understanding the Question:
We need to find the Pearson correlation coefficient, $\rho_{X,Y}$, between two random variables X and Y. We are given the distribution of X and the conditional distribution of Y given X.
Step 2: Key Formula or Approach:
The correlation coefficient is defined as: \[ \rho_{X,Y} = \frac{\text{Cov}(X, Y)}{\sigma_X \sigma_Y} \] where the covariance is given by: \[ \text{Cov}(X, Y) = E[XY] - E[X]E[Y] \] If we can show that the covariance is zero, the correlation will also be zero. We will use the law of total expectation, which states $E[Z] = E[E[Z|X]]$.
Step 3: Detailed Explanation:
First, let's calculate the necessary expected values.
- Calculate E[X]:
X follows a Uniform(-1, 1) distribution. The expected value of a uniform distribution U(a,b) is $(a+b)/2$.
\[ E[X] = \frac{-1 + 1}{2} = 0 \] - Calculate E[XY]:
We use the law of total expectation: $E[XY] = E[E[XY | X]]$.
Inside the inner expectation, X is treated as a constant: $E[XY | X=x] = x E[Y | X=x]$.
So, $E[XY | X] = X \cdot E[Y | X]$.
We need to find $E[Y|X]$. The conditional distribution of Y given X=x is Uniform($x^2 - 0.1, x^2 + 0.1$).
The expected value of this uniform distribution is the midpoint of the interval:
\[ E[Y | X=x] = \frac{(x^2 - 0.1) + (x^2 + 0.1)}{2} = \frac{2x^2}{2} = x^2 \]
Now, substitute this back: $E[XY | X] = X \cdot X^2 = X^3$.
Finally, we take the outer expectation:
\[ E[XY] = E[X^3] \]
To calculate $E[X^3]$, we integrate over the distribution of X. The PDF for U(-1,1) is $f_X(x) = 1/2$ for $x \in [-1, 1]$.
\[ E[X^3] = \int_{-1}^{1} x^3 f_X(x) dx = \int_{-1}^{1} x^3 \left(\frac{1}{2}\right) dx \]
Since the integrand $x^3$ is an odd function and the interval of integration [-1, 1] is symmetric about 0, the integral is 0.
\[ E[XY] = 0 \]
- Calculate Cov(X, Y):
\[ \text{Cov}(X, Y) = E[XY] - E[X]E[Y] \]
\[ \text{Cov}(X, Y) = 0 - (0) \times E[Y] = 0 \]
(We don't even need to calculate E[Y], but we know Cov(X,Y) is 0).
- Calculate Correlation $\rho_{X,Y$:}
Since the covariance in the numerator is 0, the entire correlation coefficient is 0 (as long as the variances are non-zero, which they are).
\[ \rho_{X,Y} = \frac{0}{\sigma_X \sigma_Y} = 0 \]
Step 4: Final Answer:
The value of the correlation between X and Y is 0.