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Suppose that X is a Poisson random variable with parameter 5. Let Y = min(X, 5). Determine the following:

(a) The probability mass function of X (
b) The probability mass function of Y
(c) E[Y]

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Final answer:

The probability mass function (PMF) of a Poisson random variable X with parameter λ is given by P(X = k) = (λ^k * e^(-λ))/k!. To find the PMF of Y = min(X, 5), calculate the probabilities for each possible value of Y. The expected value of Y, E[Y], is the weighted average of the possible values of Y, weighted by their probabilities.

Step-by-step explanation:

Probability Mass Function of X:


The probability mass function (PMF) of a Poisson random variable X with parameter λ is given by

P(X = k) = (λ^k * e^(-λ))/k!


For X with parameter 5, the PMF of X is

P(X = k) = (5^k * e^(-5))/k!



Probability Mass Function of Y:


Let Y = min(X, 5).
To find the PMF of Y, we need to find the probability that Y takes on each possible value.
Since Y is the minimum of X and 5, Y can take on the values 0, 1, 2, 3, 4, and 5.

P(Y = 0) = P(X = 0) = (5^0 * e^(-5))/0! = e^(-5)

P(Y = 1) = P(X = 1) = (5^1 * e^(-5))/1! = (5 * e^(-5))

P(Y = 2) = P(X = 2) = (5^2 * e^(-5))/2! = (25 * e^(-5))/2

P(Y = 3) = P(X = 3) = (5^3 * e^(-5))/3! = (125 * e^(-5))/6

P(Y = 4) = P(X = 4) = (5^4 * e^(-5))/4! = (625 * e^(-5))/24

P(Y = 5) = 1 - P(Y > 5) = 1 - P(X > 5) = 1 - (P(X = 6) + P(X = 7) + ...)

Expected Value of Y:


To find the expected value of Y, we need to calculate the weighted average of the possible values of Y, weighted by their probabilities.
The expected value of Y, denoted as E[Y], is given by E[Y] = Σ(y * P(Y = y)) for all possible values of Y.

E[Y] = 0 * P(Y = 0) + 1 * P(Y = 1) + 2 * P(Y = 2) + 3 * P(Y = 3) + 4 * P(Y = 4) + 5 * P(Y = 5)

Substituting the probabilities from above, we get E[Y] = 0 * e^(-5) + 1 * (5 * e^(-5)) + 2 * (25 * e^(-5))/2 + 3 * (125 * e^(-5))/6 + 4 * (625 * e^(-5))/24 + 5 * P(Y = 5)

Simplifying further, we get E[Y] = (1/120) * (5 * e^(-5) + 50 * e^(-5) + 125 * e^(-5) + 125 * e^(-5) + 625 * e^(-5) + 5)

Therefore, E[Y] = (1/120) * (930 * e^(-5) + 5

SPJ1

User Martinho
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