Question

3. Let X be random variable with probability density function x(x)4 for 0 x 1, (Note: fx (x) = 0 outside this domain.) (a) Find E[X] and Var[X] (b) Let Y- X2 +5. Find E[Y] and Var[Y]. (c) Find PX 112 ).

Answer 1

3.

(a)

$E[X]=\int_{0}^{1}xf(x)dx =\int_{0}^{1}x (4x^3)dx =\int_{0}^{1}4x^4dx =4\left [ \frac{x^5}{5} \right ]_0^1$

$=4\left [ \frac{1^5}{5}-\frac{0^5}{5} \right ] = \frac{4}{5} = 0.8$

$Var[X] = E[X^2] - ( E[X])^2$

$E[X^2]=\int_{0}^{1}x^2f(x)dx =\int_{0}^{1}x^2 (4x^3)dx =\int_{0}^{1}4x^5dx =4\left [ \frac{x^6}{6} \right ]_0^1$

$=4\left [ \frac{1^6}{6}-\frac{0^6}{6} \right ] = \frac{4}{6}$

$Var[X] = \frac{4}{6}- \left ( \frac{4}{5} \right )^2 =$$Var[X] = \frac{4}{6}- \left ( \frac{4}{5} \right )^2 = 0.0267$

(b)

We apply transformation :

$Y = X^2 + 5$

$=> X = \sqrt{Y-5}$

$Jacobian = |J| =\left | \frac{\mathrm{d} x }{\mathrm{d} y} \right | = \left | \frac{\mathrm{d} \sqrt{y-5} }{\mathrm{d} y} \right | = \frac{1}{2}(y-5)^{1/2-1}$

$= \frac{1}{2}(y-5)^{-1/2}$

So, the density function of Y will be :

$f(y) = 4x^3.|J| = 4(\sqrt{y-5})^3. \frac{1}{2}(y-5)^{-1/2} = 2(y-5)^{3/2-1/2} = 2(y-5)$

Since,

$0\leq x\leq 1$

$=>0\leq \sqrt{y-5}\leq 1$

$=>5\leq y\leq 6$

$E[Y]=\int_{5}^{6}yf(y)dy =\int_{5}^{6}y (2(y-5))dy =\int_{5}^{6}2(y^2-5y)dy =2\left [ \frac{y^3}{3}- 5\frac{y^2}{2}\right ]_5^6$

$=2\left [ \frac{6^3}{3}- 5\frac{6^2}{2}-\left ( \frac{5^3}{3}- 5\frac{5^2}{2} \right )\right ]$

$=5.667$

$Var[Y] = E[Y^2] - ( E[Y])^2$

$E[Y^2]=\int_{5}^{6}y^2f(y)dy =\int_{5}^{6}y^2 (2(y-5))dy =\int_{5}^{6}2(y^3-5y^2)dy =2\left [ \frac{y^4}{4}- 5\frac{y^3}{3}\right ]_5^6$

$=2\left [ \frac{6^4}{4}- 5\frac{6^3}{3}-\left ( \frac{5^4}{4}- 5\frac{5^3}{3} \right )\right ]$

$=32.1667$

$Var[Y] = E[Y^2] - ( E[Y])^2 = 32.1667 - (5.667)^2 = 0.0518$

(c)

$P(X\leq 2/3 | X\geq 1/3) =\frac{P(X\leq 2/3,X\geq 1/3)}{P(X\geq 1/3)}$

$=\frac{P(1/3\leq X\leq 2/3)}{P(X\geq 1/3)}$

$P(1/3\leq X\leq 2/3) = \int_{1/3}^{2/3}f(x)dx = \int_{1/3}^{2/3}4x^3dx = 4\left [ \frac{x^4}{4} \right ]_{1/3}^{2/3}$

$= 4\left [ \frac{(2/3)^4}{4}-\frac{(1/3)^4}{4} \right ]$

$=0.185$

$P(X\geq 1/3) =\int_{1/3}^{1}f(x)dx = \int_{1/3}^{1}4x^3dx = 4\left [ \frac{x^4}{4} \right ]_{1/3}^{1}$

$= 4\left [ \frac{(1)^4}{4}-\frac{(1/3)^4}{4} \right ]$

$=0.988$

$P(X\leq 2/3 | X\geq 1/3) =\frac{P(X\leq 2/3,X\geq 1/3)}{P(X\geq 1/3)} = \frac{0.185}{0.988} = 0.1875$