Let y be a random variable with p(y) given in the accompanying table. find e(y ), e(1/y ), e(y 2 − 1), and v(y ). y 1 2 3 4 p(y)
.4 .3 .2 .1
1 answer:
<span>E[Y] = 0.4·1 + 0.3·2 + 0.2·3 + 0.1·4 = 2
E[1/Y] =0.4·1/1 + 0.3·1/2 + 0.2·1/3 + 0.1·1/4 = 0.4 + 0.15 + 0.0666 + 0.025?0.64
V[Y] =E[Y2]-E[Y]2= (0.4)·12+(0.3)·22+(0.2)·32+(0.1)·42-22= 0.4+1.2+1.8+1.6-4= 5-4 = 1</span>
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