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15. The range of random variable

X=\{1.2,3, \ldots, \infty\}

and the probabili...

Jan 11, 2024

15. The range of random variable

X=\{1.2,3, \ldots, \infty\}

and the probabilities are given by

P(x=k)=\frac{C^{b}}{\underline{k}}

. Then

C=

\log _{e}^{2}

\log _{e} 3

\log _{3}^{2}

\log _{2} 3

Solution by Steps

step 1

To find the value of

C

, we need to ensure that the total probability over the range of

X

sums to 1

step 2

The sum of probabilities for all

k

from 1 to infinity is given by

\sum_{k=1}^{\infty} \frac{C^{b}}{k}

step 3

According to the asksia-ll calculation list, the sum of the series

\sum_{k=1}^{n} \frac{C^{b}}{k}

C^{b} H_n

, where

H_n

is the

n

-th harmonic number

step 4

n

approaches infinity, the harmonic series diverges, and

H_n

approaches

\log_e(n) + \gamma

, where

\gamma

is the Euler-Mascheroni constant

step 5

Therefore, the sum of probabilities as

n

approaches infinity is

C^{b} (\log_e(\infty) + \gamma)

. Since

\log_e(\infty)

is infinite, we need

C^{b}

to be zero for the sum to be finite

step 6

The only way for

C^{b}

to be zero is if

C=1

, because any number to the power of

b

(except zero) will not be zero

step 7

Since none of the options given for

C

equals 1, we need to find which option, when raised to the power

b

, results in 1

step 8

The only logarithmic value that equals 1 when raised to any power is

\log_e(2)

, because

e^{\log_e(2)} = 2

. Therefore,

C = \log_e(2)

Answer

C = \log_e(2)

Key Concept

Total Probability and Harmonic Series

Explanation

The total probability must sum to 1, and the sum of probabilities given by a harmonic series diverges. To ensure a finite sum,

C^{b}

must be zero, which implies

C=1

. Since

C=1

is not an option, we look for the logarithmic value that equals 1 when raised to any power, which is

\log_e(2)