# If X ∼ N (0, 1) and Y = X2 then the correlation coefficient r (X, Y) is

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

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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)μx = 0σx2 = 1 ⇒ σx = 1;

And the probability density function will be

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

μx = 0 ⇒ E(X) = 0;

var(X) = σX2 = 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 = σy2 = 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;

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