If a random variable X following Poisson distribution has expectation λ, then which of the following statements is true?

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SSC CGL Tier-II (JSO) 2022 Official Paper (Held On: 4 March 2023)
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  1. \(\rm P(X=\lambda)=\frac{\lambda}{\lambda+1}P(X=\lambda+1)\)
  2. \(\rm P(X=\lambda)=\frac{\lambda+1}{\lambda}P(X=\lambda+1)\)
  3. \(\rm P(X=\lambda)=\frac{\lambda}{(\lambda+1)e}P(X=\lambda+1)\)
  4. \(\rm P(X=\lambda)=\frac{\lambda e}{\lambda+1}P(X=\lambda+1)\)

Answer (Detailed Solution Below)

Option 2 : \(\rm P(X=\lambda)=\frac{\lambda+1}{\lambda}P(X=\lambda+1)\)
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Key Points A random variable X is given that follows a Poisson distribution with parameter λ, which represents the expected number of events in an interval. The probability mass function (PMF) of a Poisson-distributed random variable is given as follows: \(P(X = k) = λ^k e^{(-λ)} {1 \over k!} \)for k = 0, 1, 2, ...
Therefore, we can write: \(P(X = λ) = λ^λ e^{(-λ)} {1 \over λ!}\)and \(P(X = λ + 1) = λ^{(λ + 1)} e^{{(-λ)} \over (λ + 1)!}\)
Taking the ratio of these two probabilities, we have: 
\({P(X = \lambda) \over P(X = \lambda + 1)} = {{{[\lambda^\lambda \times e^{(-\lambda)} {1 \over \lambda!}]} \over {[\lambda^{(\lambda + 1)} \times e^{(-\lambda)} {1 \over (\lambda + 1)!}]}}}\)
\(={{\lambda^\lambda \over {\lambda^{(\lambda+1)}}} \times {{(\lambda+1)!} \over λ!}}\)
\(= {\lambda +1 \over \lambda}\)

Rearranging the above equation gives us the relation between the two probabilities: 
\(\rm P(X=\lambda)=\frac{\lambda +1}{\lambda}P(X=\lambda+1)\)
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