antiderivative of k/x^{k+1} = -1/x^k So this defines a pdf as the integral between theta and infinity is 1. The cdf is F(x) = 1-(theta/x)^k.
P(a \leq x \leq b) = F(b) - F(a)
E(X) = k theta /(k-1) with E(X) = infinity
Oliver Q.
asked 04/02/20A random variable X has the pdf f(x) = kθ^k/ x^(k+1) , x ≥ θ,where k and θ are parameters, and > 0. This random variable is called Pareto random variable.
(a) Sketch the graph of f(x).
(b) Show that this pdf is valid.
(c) Derive the cdf of X.
(d) For θ < a < b, obtain an expression for the probability P(a ≤ X ≤ b).
(e) Compute E(X) when k > 1.
(f) What can you say about E(X) if k = 1?
antiderivative of k/x^{k+1} = -1/x^k So this defines a pdf as the integral between theta and infinity is 1. The cdf is F(x) = 1-(theta/x)^k.
P(a \leq x \leq b) = F(b) - F(a)
E(X) = k theta /(k-1) with E(X) = infinity
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