Find the moment generating of the following distribution P(X=x)=qp, x=0,1,2
ودة Hence find the mean and variances
To find the moment generating function of the distribution, we can use the formula for the moment generating function of a discrete random variable:
M(t) = E(e^(tX)) = Σ e^(tx) * P(X=x)
Given that P(X=x) = qp for x=0,1,2, we can plug in these values into the formula:
M(t) = e^(t*0) * qp + e^(t*1) * qp + e^(t*2) * qp
M(t) = q + qe^t + qe^(2t)
M(t) = q(1 + e^t + e^(2t))
Next, to find the mean and variance of the distribution, we can differentiate the moment generating function with respect to t and evaluate at t=0 for the mean, and t=0 for the variance.
Mean (μ) = M'(0)
μ = q(0 + e^0 + 2e^(0))
μ = q(1 + 2)
μ = 3q
Variance (σ^2) = E(X^2) - [E(X)]^2
σ^2 = M''(0) - [M'(0)]^2
σ^2 = q(0 + 0 + 4e^(0)) - (3q)^2
σ^2 = 4q - 9q^2
Therefore, the moment generating function of the distribution is M(t) = q(1 + e^t + e^(2t)), the mean is 3q, and the variance is 4q - 9q^2.