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

a) The value of A = 2

b) The value of
`B = (1)/(12)`

Explanation:

a)

Given that:

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

The probability function;
`P[X = n] = (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) (1)/(3^n)= 1`

`A * ((1)/(3))/(1 - (1)/(3)) = 1`

`A * ((1)/(3))/((2)/(3)) = 1`

`A * (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 [(x^3)/(3) +x \Big ]^3_0 = 1`

`B \Big [(3^3)/(3) +3 \Big ]= 1`

`B \Big [(27)/(3) +3 \Big ] = 1`

B [ 9 +3 ] = 1

B [ 12 ] = 1

Divide both sides by 12

`B = (1)/(12)`