Correct Answer - Option 2 : zero

__Concept:__

The **correlation coefficient** is given by

**\(r = \frac{{cov\left( {X,Y} \right)}}{{{\sigma _x}{\sigma _y}}}\)**

**Cov(X,Y) = E(XY) – E(X) E(Y)**

Where **cov(X, Y) = covariance of X and Y**; **σ = standard deviation**; **E = expectation**;

__Calculation:__

Given **X ∼ N (0, 1)** ⇒ **μ**_{x} = 0 & **σ**_{x}^{2} = 1 ⇒ σ_{x} = 1;

And the **probability density function** will be

\(f\left( x \right) = \frac{1}{{\sqrt {2\pi } }}{e^{ - \frac{{{x^2}}}{2}}};\; - \infty < x < + \infty \)

μ_{x} = 0** ⇒ E(X) = 0**;

var(X) = σ_{X}^{2}** **=** E (X**^{2}) – [E (X)]^{2} **= 1 ⇒ E (X**^{2}) = 1;

Given **Y = X**^{2}; therefore **Y is a chi-square distribution with 1 degree of freedom.**

For a **chi-square distribution with degree of freedom (n) = 1,**

Expectation =** E(Y) = n = 1**; variance =** σ**_{y}^{2} = 2 × n = 2;

Now

**E(XY) = E (X**^{3}) = \(\int\limits_{ - \infty }^{ + \infty } {{x^3}{e^{( - {x^2}/2)}}} dx\)

Hence, the **function inside the integral is odd**, we can say it is **symmetric w.r.t zero.**

⇒ **E(X**^{3}) = 0;

Now

Cov(X, X^{2}) = E (X^{3}) – E(X) E(X^{2}) = 0 – 0 × 1 = **0**;

\(\therefore r = \frac{0}{{1 \times \sqrt 2 }} = 0\)