To illustrate the proof of Theorem 1, consider the ran-dom variable X, which takes on the values −2, −1, 0, 1, 2, and 3 with probabilities f(−2), f(−1), f(0), f(1), f(2),and f(3). If g(X) = X2, find(a) g1, g2, g3, and g4, the four possible values of g(x);(b) the probabilities P[g(X) = gi] for i = 1, 2, 3, 4;(c) E[g(X)] = 4i=1gi ·P[g(X) = gi], and show that it equals xg(x)·f(x)
To illustrate the proof of Theorem 1, consider the ran-dom variable X, which takes on the values −2, −1, 0, 1, 2, and 3 with probabilities f(−2), f(−1), f(0), f(1), f(2),and f(3). If g(X) = X2, find(a) g1, g2, g3, and g4, the four possible values of g(x);(b) the probabilities P[g(X) = gi] for i = 1, 2, 3, 4;(c) E[g(X)] = 4i=1gi ·P[g(X) = gi], and show that it equals xg(x)·f(x)
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.8: Probability
Problem 23E
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Question
To illustrate the proof of Theorem 1, consider the ran-
dom variable X, which takes on the values −2, −1, 0, 1,
dom variable X, which takes on the values −2, −1, 0, 1,
2, and 3 with probabilities f(−2), f(−1), f(0), f(1), f(2),
and f(3). If g(X) = X2, find
(a) g1, g2, g3, and g4, the four possible values of g(x);
(b) the probabilities P[g(X) = gi] for i = 1, 2, 3, 4;
(c) E[g(X)] =
4
i=1
gi ·P[g(X) = gi], and show that
and f(3). If g(X) = X2, find
(a) g1, g2, g3, and g4, the four possible values of g(x);
(b) the probabilities P[g(X) = gi] for i = 1, 2, 3, 4;
(c) E[g(X)] =
4
i=1
gi ·P[g(X) = gi], and show that
it equals
x
g(x)·f(x)
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