Question

7.4 Let X ~U(_ 1,1 ) and Y =X2. What are the density, the distribution function, the mean, and the variance of Y What is Pr[Y s0.5]? a. b.

Answer 1

Sol:

Here we are given the distribution as:

$X \sim U[-1, 1]$

a) Here, the CDF for X is computed as:

$F_x(x) = \frac{x + 1}{2}$

Now the CDF for Y = X² is computed as:

$P(Y \leq y) = P(X^2 \leq y) = P(X \leq \sqrt{y})$

Now using the CDF for X, we get:

$P(Y \leq y) = P(X^2 \leq y) = P(X \leq \sqrt{y}) = \frac{\sqrt{y}+1}{2}$

This is the CDF for Y here. This is the required distributive function for Y.

Now the PDF for Y is computed by differentiating the CDF with respect to y as:

$f_y(y) =\frac{\mathrm{d} ( \frac{\sqrt{y}+1}{2})}{\mathrm{d} y} = \frac{1}{4\sqrt{y}}$

This is the required density function for Y.

Now the mean of Y here is computed as:

$E(Y) = \int_{0}^{1}yf(y)dy = \int_{0}^{1}\frac{\sqrt{y}}{4} dy = \left |\frac{y^{3/2}}{4*(3/2)} \right |_{0}^{1} = \frac{1}{6}$

Therefore 1/6 = 0.1667 is the mean of Y here.

The second moment of Y here is computed as:

$E(Y^2) = \int_{0}^{1}y^2f(y)dy = \int_{0}^{1}\frac{y^{1.5}}{4} dy = \left |\frac{y^{2.5}}{4*2.5} \right |_{0}^{1} = \frac{1}{10}$

Therefore the variance of Y here is computed as:

$Var(Y) = E(Y^2) - (E(Y))^2 = \frac{1}{10} - \frac{1}{6^2} = 0.0722$

Therefore 0.0722 is the variance of Y here.

b) The required probability here is computed as:

$P(Y \leq 0.5) = \int_{0}^{0.5} f(y) dy = \int_{0}^{0.5} \frac{1}{4\sqrt{y}} dy = \left [\frac{\sqrt{y}}{2} \right ]_{0}^{0.5} = \frac{\sqrt{0.5}}{2} = 0.3536$

Therefore 0.3536 is the required probability here.