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 (x2 - 0.1, x2 + 0.1) dist. The value of correlation (X, Y) is______ .

The correct option is 0.

To find the correlation between the random variables $X$ and $Y$, we use the formula for correlation:

$$\mathrm{Corr}(X,Y)=\frac{\mathrm{Cov}(X,Y)}{{\sigma }_X{\sigma }_Y}$$

where the covariance is defined as:

$$\mathrm{Cov}(X,Y)=E\left[XY\right]-E\left[X\right]E\left[Y\right]$$

First, we find the expected value of $X$. Since $X$ is uniformly distributed on the interval $(-1,1)$, its probability density function is symmetric around 0. Thus, its expected value is:

$$E\left[X\right]=0$$

Next, we calculate the expected value of the product $XY$ by applying the law of total expectation:

$$E\left[XY\right]=E\left[E\right[XY\left|X\right]]=E[X·E\left[Y\right|X\left]\right]$$

We are given that the conditional distribution of $Y$ given $X=x$ is $\mathrm{Uniform}(x^2-0.1,x^2+0.1)$. The mean of a uniform distribution over $[a,b]$ is given by $\frac{a+b}{2}$. Therefore, the conditional expectation of $Y$ given $X=x$ is:

$$E\left[Y\right|X=x]=\frac{(x^2-0.1)+(x^2+0.1)}{2}=x^2$$

Substituting this back into the expectation of $XY$:

$$E\left[XY\right]=E[X·X^2]=E\left[X^3\right]$$

Because the distribution of $X$ is symmetric about 0, the expectation of any odd power of $X$ over this symmetric interval is 0. We can show this explicitly by integrating over the probability density function $f\left(x\right)=\frac{1}{2}$ for $x\in (-1,1)$:

$$E\left[X^3\right]={\int }_{-1}^1{x}^3·\frac{1}{2}dx={\left[\frac{x^4}{8}\right]}_{-1}^1=\frac{1}{8}-\frac{1}{8}=0$$

This gives us $E\left[XY\right]=0$.

Now we substitute our values back into the covariance formula:

$$\mathrm{Cov}(X,Y)=E\left[XY\right]-E\left[X\right]E\left[Y\right]=0-0·E\left[Y\right]=0$$

Since the covariance between $X$ and $Y$ is 0, the correlation is also 0:

$$\mathrm{Corr}(X,Y)=0$$