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iris [78.8K]
3 years ago
5

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.
Mathematics
1 answer:
kramer3 years ago
5 0

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}

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