Question

If X ∼ N (0, 1) and Y = X² then the correlation coefficient r (X, Y) is
1. one
2. zero
3. infinity
4. two

Answer 1

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_x² = 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_X² = E (X²) – [E (X)]² = 1 ⇒ E (X²) = 1; 

Given Y = X²; 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_y² = 2 × n = 2;

Now

E(XY) = E (X³) = \(\int\limits_{ - \infty }^{ + \infty } {{x^3}{e^{( - {x^2}/2)}}} dx\)

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

⇒ E(X³) = 0;

Now  

Cov(X, X²) = E (X³) – E(X) E(X²) = 0 – 0 × 1 = 0;

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