For any positive real number t, defined by a function P(X = x) = e–t (1 – e–t)x – 1 represent a probability function of a random variable X assuming the values 1, 2, 3, …, what is the variance of X of this distribution ?
et (et – 1)
The given probability function P(X = x) = e–t (1 – e–t)x – 1, x = 1, 2, 3, …, has the form qx − 1p, so X follows a geometric distribution (the trial on which the first success occurs).
Its parameters are p = e–t (success probability) and q = 1 – e–t; for t > 0 both lie between 0 and 1, so this is a valid probability function.
For this geometric distribution, E(X) = 1/p = et and Var(X) = q/p2.
Substituting: \(Var(X) = \frac{1 - e^{-t}}{e^{-2t}} = (1 - e^{-t})e^{2t} = e^{2t} - e^{t} = e^{t}(e^{t} - 1)\).
Hence, the answer is et (et – 1).
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.................... is a client server application that identifies each host on the internet with a unique user-friendly name.
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(i) Kernel
(ii) Interface
(iii) Nucleus
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(i) Machine language
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(iii) High level language
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