If for a Poisson variate x,P(x=0)=P(x=1) = k, prove that k=1/e. Lesson name Poisson distribution
To prove that k = 1/e for a Poisson variate x where P(x=0) = P(x=1) = k, we need to use the properties of the Poisson distribution.
The Poisson distribution is commonly used to model the number of events occurring in a fixed interval of time or space, given the average rate of occurrence. The probability mass function (PMF) of the Poisson distribution is given by:
P(x) = (e^(-λ) * λ^x) / x!
Where λ is the average rate of occurrence.
Now, we are given that P(x=0) = P(x=1) = k. Let's use this information to find the value of k.
1. P(x=0) = k
Substituting x=0 in the Poisson PMF, we have:
(e^(-λ) * λ^0) / 0! = k
Simplifying, we get:
e^(-λ) = k ---(equation 1)
2. P(x=1) = k
Substituting x=1 in the Poisson PMF, we have:
(e^(-λ) * λ^1) / 1! = k
Simplifying, we get:
λ * e^(-λ) = k ---(equation 2)
To prove that k = 1/e, we need to equate equations 1 and 2 and solve for k.
Using equation 1, we can rewrite the equation as:
λ * e^(-λ) = e^(-λ)
λ = 1
Therefore, we have established that λ (the average rate of occurrence) is equal to 1.
Now, substituting λ=1 in equation 1, we get:
e^(-1) = k
The value of e^(-1) is approximately 0.36788.
Therefore, k = 0.36788, which is equal to 1/e.
Hence, we have proven that k = 1/e for a Poisson variate x where P(x=0) = P(x=1).