Let X, N, Y and Z be random variables. The variables X and N are independent of each other. X is uniformly distributed between -1 and 1; N follows Normal distribution with zero mean and unity variance. Y and Z are defined as, Y = X +N and Z = X2 + N. Which of the following pairs represents the values of correlation between X and Y and that between X and Z?

The correct option is 1/3 and 0.

Note: In the context of this question, "correlation" refers to the covariance between the random variables.

1. Understanding the given distributions:
We are given two independent random variables, $X$ and $N$.
• $X$ is uniformly distributed between -1 and 1. The probability density function of $X$ is symmetric around its mean:
$$E\left[X\right]=0$$
The variance of a continuous uniform distribution on the interval $[a,b]$ is given by:
$$\mathrm{Var}\left(X\right)=\frac{{(b-a)}^2}{12}$$
Substituting $a=-1$ and $b=1$:
$$\mathrm{Var}\left(X\right)=\frac{{(1-(-1\left)\right)}^2}{12}=\frac{2^2}{12}=\frac{4}{12}=\frac{1}{3}$$
• $N$ follows a normal distribution with zero mean and unity variance:
$$E\left[N\right]=0,\mathrm{Var}\left(N\right)=1$$
• Since $X$ and $N$ are independent, their covariance is zero:
$$\mathrm{Cov}(X,N)=0$$

2. Finding the covariance (correlation) between X and Y:
The variable $Y$ is defined as $Y=X+N$.
Using the properties of covariance, we get:
$$\mathrm{Cov}(X,Y)=\mathrm{Cov}(X,X+N)$$
By linearity of covariance:
$$\mathrm{Cov}(X,Y)=\mathrm{Cov}(X,X)+\mathrm{Cov}(X,N)$$
Since $\mathrm{Cov}(X,X)=\mathrm{Var}\left(X\right)$ and $\mathrm{Cov}(X,N)=0$:
$$\mathrm{Cov}(X,Y)=\mathrm{Var}\left(X\right)+0=\frac{1}{3}$$

3. Finding the covariance (correlation) between X and Z:
The variable $Z$ is defined as $Z=X^2+N$.
We compute the covariance as:
$$\mathrm{Cov}(X,Z)=\mathrm{Cov}(X,X^2+N)$$
Using linearity of covariance:
$$\mathrm{Cov}(X,Z)=\mathrm{Cov}(X,X^2)+\mathrm{Cov}(X,N)$$
Since $X$ and $N$ are independent, $\mathrm{Cov}(X,N)=0$:
$$\mathrm{Cov}(X,Z)=\mathrm{Cov}(X,X^2)$$
By definition of covariance:
$$\mathrm{Cov}(X,X^2)=E[X\cdot X^2]-E\left[X\right]E\left[X^2\right]=E\left[X^3\right]-E\left[X\right]E\left[X^2\right]$$
We know that $E\left[X\right]=0$. Next, we evaluate the third moment $E\left[X^3\right]$:
$$E\left[X^3\right]={\int }_{-1}^1{x}^3\cdot f\left(x\right)dx$$
Since $x^3$ is an odd function and the probability density function $f\left(x\right)=\frac{1}{2}$ is symmetric over the interval $[-1,1]$, the integral of the odd function over this symmetric interval is zero:
$$E\left[X^3\right]=0$$
Substituting these values back into the covariance equation:
$$\mathrm{Cov}(X,Z)=0-0\cdot E\left[X^2\right]=0$$

Therefore, the value of the covariance (correlation) between $X$ and $Y$ is 1/3, and between $X$ and $Z$ is 0.