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If the random variable X follows the Poisson distribution such that P[X = k] = P[X = k+1], then the mean value of X is:

Solution

Correct Option: 4

For a Poisson distribution where P[X=k]=P[X=k+1]P[X = k] = P[X = k+1], the mean value of XX needs to be determined.

The probability formula for a Poisson distribution is:

P[X=n]=λn×eλn!P[X = n] = \dfrac{\lambda^n \times e^{-\lambda}}{n!}

where λ\lambda is the mean of the distribution.


Given that P[X=k]=P[X=k+1]P[X = k] = P[X = k+1]:

P[X=k]=λk×eλk!P[X = k] = \dfrac{\lambda^k \times e^{-\lambda}}{k!}

P[X=k+1]=λk+1×eλ(k+1)!P[X = k+1] = \dfrac{\lambda^{k+1} \times e^{-\lambda}}{(k+1)!}


Setting them equal:

λk×eλk!=λk+1×eλ(k+1)!\dfrac{\lambda^k \times e^{-\lambda}}{k!} = \dfrac{\lambda^{k+1} \times e^{-\lambda}}{(k+1)!}

Canceling eλe^{-\lambda} from both sides:

λkk!=λk+1(k+1)!\dfrac{\lambda^k}{k!} = \dfrac{\lambda^{k+1}}{(k+1)!}


Since (k+1)!=(k+1)×k!(k+1)! = (k+1) \times k!:

λkk!=λk+1(k+1)×k!\dfrac{\lambda^k}{k!} = \dfrac{\lambda^{k+1}}{(k+1) \times k!}

Canceling k!k!:

λk=λk+1k+1\lambda^k = \dfrac{\lambda^{k+1}}{k+1}


(k+1)×λk=λk+1(k+1) \times \lambda^k = \lambda^{k+1}

(k+1)×λk=λ×λk(k+1) \times \lambda^k = \lambda \times \lambda^k

Dividing both sides by λk\lambda^k:

k+1=λk + 1 = \lambda


Therefore, the mean value λ=k+1\lambda = k + 1

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