STA301 ASSIGNMEN NO. 2 Spring 2022 Solution

STA301

ASSIGNMEN NO. 2

Spring 2022

Due Date: 30 – 8 - 2022

QUESTION NO. 1:-

Let X be a continuous random variable whose probability density function is:

$f\left( x \right) = \frac{{{x^3}}}{4}$

SOLUTION:-

Given that:-

$f\left( x \right) = \frac{{{x^3}}}{4}$

The function $f\left( x \right)$ will be a density function, if:

(i) $f\left( x \right) > 0\,\,for\,\,every\,\,x.$

(ii) $\int\limits_{ - \infty }^\infty {f\left( x \right)dx = 1}$

The first condition is satisfied when, x > 0.

$\int\limits_{ - \infty }^\infty {f\left( x \right)dx = 1}$

$\int\limits_0^c {\frac{{{x^3}}}{4}dx = 1}$

$\frac{1}{4}\int\limits_0^c {{x^3}dx = 1}$

$\frac{1}{4}\left[ {\left| {\frac{{{x^{3 + 1}}}}{{3 + 1}}} \right|_0^c} \right] = 1$

$\frac{1}{4}\left[ {\left| {\frac{{{x^4}}}{4}} \right|_0^c} \right] = 1$

$\frac{1}{4}\left[ {\frac{{{{\left( c \right)}^4}}}{4} - \frac{{{{\left( 0 \right)}^4}}}{4}} \right] = 1$

$\frac{1}{4}\left[ {\frac{{{c^4}}}{4}} \right] = 1$

$\frac{{{c^4}}}{{16}} = 1$

${c^4} = 16$

${c^4} = {2^4}$

c = 2

QUESTION NO. 2:-

From the following joint p.d of X and Y. Find E(X), Var(Y) and Cov(X,Y).

X	Y	0	1	2	3	g(X)
0		0	0.25	0.05	0.05	0.35
1		0.03	0.10	0.10	0.15	0.38
2		0.07	0.10	0	0.10	0.27
h(y)		0.10	0.45	0.15	0.30	1

SOLUTION:-

Given that:

X	Y	0	1	2	3	g(X)
0		0	0.25	0.05	0.05	0.35
1		0.03	0.10	0.10	0.15	0.38
2		0.07	0.10	0	0.10	0.27
h(y)		0.10	0.45	0.15	0.30	1

E(X) =?

Var(Y) =?

Cov(X,Y) =?

Now,

We know that:-

Formula:

$E\left( X \right) = \sum {Xig\left( {Xi} \right)}$

$= \left( {0 \times 0.35} \right) + \left( {1 \times 0.38} \right) + \left( {2 \times 0.27} \right)$

= 0 + 0.38 + 0.54

$E\left( X \right)\,\,\,\,\,\,\, = \,\,0.92$

Now,

We know that:-

Formula:

$E\left( Y \right) = \sum {Yjh\left( {Yj} \right)}$

$= \left( {0 \times 0.10} \right) + \left( {1 \times 0.45} \right) + \left( {2 \times 0.15} \right) + \left( {3 \times 0.30} \right)$

= 0 + 0.45 + 0.30 + 0.90

$E\left( Y \right) = 1.65$

Now,

We know that:-

Formula:

$E\left( {{Y^2}} \right) = \sum {{Y^2}jh\left( {Yj} \right)}$

$= \left( {0 \times 0.10} \right) + \left( {1 \times 0.45} \right) + \left( {4 \times 0.15} \right) + \left( {9 \times 0.30} \right)$

= 0 + 0.45 + 0.60 + 2.70

$E\left( {{Y^2}} \right) = 3.75$

Now,

We know that:-

Formula:

$Var\left( Y \right) = \left[ {E\left( {{Y^2}} \right)} \right] - {\left[ {E\left( Y \right)} \right]^2}$

$= \left( {3.75} \right) - {\left( {1.65} \right)^2}$

= 3.75 - 2.7225

$Var\left( Y \right) = 1.0275$

And

We know that:-

Formula:

$E\left( {XY} \right) = \sum {\sum {\left( {{x_i}{y_j}} \right)f\left( {{x_i}{y_j}} \right)} }$

$E\left( {XY} \right) =$ [ ${\left( {0 \times 0 \times 0} \right) + \left( {0 \times 1 \times 0.25} \right) + \left( {0 \times 2 \times 0.05} \right) + }$ ${\left( {0 \times 3 \times 0.05} \right) + \left( {1 \times 0 \times 0.03} \right)}$ ${\left( {1 \times 1 \times 0.10} \right)}$

${\left( {1 \times 2 \times 0.10} \right)}$ ${ + \left( {1 \times 3 \times 0.15} \right) + \left( {2 \times 0 \times 0.07} \right) + \left( {2 \times 1 \times 0.10} \right)}$ ${ + \left( {2 \times 2 \times 0} \right) + \left( {2 \times 3 \times 0.10} \right)}$ ]