Question

A) Let X be a random variable that can assume only positive integer values, and assume its probability function is P(X -n) A/3^n for some constant A (n> 1). Find A.
b) Let X be a continuous random variable that can assume values between 0 and 3, and assume its density function is fx(x)- B(x2 +1) with some constant B (0< x ,3). Find B.

Answer 1

Answer:

a) The value of A = 2

b) The value of $B = \dfrac{1}{12}$

Step-by-step explanation:

a)

Given that:

X should be the random variable that assumes only positive integer values.

The probability function; $P[X = n] = \dfrac{A}{3^n}$ for some constant A and n ≥ 1.

Then, let $\sum \limits ^{\infty}_{n =1} P[X =n] = 1$

This implies that:

$A \sum \limits ^{\infty}_{n =1} \dfrac{1}{3^n}= 1$

$A \times \dfrac{\dfrac{1}{3}}{1 - \dfrac{1}{3}} = 1$

$A \times \dfrac{\dfrac{1}{3}}{\dfrac{2}{3}} = 1$

$A \times \dfrac{1}{2}=1$

A = 2

Thus, the value of A = 2

b)

Suppose X represents a e constant A (n> 1). Find A.

b) Let X be a continuous random variable that can assume values between 0 and 3

Then, the density function of x is:

$f_x(x) = \left \{ {{B(x^2+1)} \ \ \ 0 \le x \le 3 \ \ \ \atop {0} \ \ \ otherwise} \right.$

where; B is constant.

Then, using the property of the probability density function:

$\int ^3_0 \ B (x^2+1 ) \ dx = 1$

Taking the integral, we have:

$B \Big [\dfrac{x^3}{3} +x \Big ]^3_0 = 1$

$B \Big [\dfrac{3^3}{3} +3 \Big ]= 1$

$B \Big [\dfrac{27}{3} +3 \Big ] = 1$

B [ 9 +3 ] = 1

B [ 12 ] = 1

Divide both sides by 12

$B = \dfrac{1}{12}$