Let Y X2 + 1. Compute E[Y] and sigma 2 y if Solution P(Y>t) = P(X1>t,X2>t,X3>t)

Question

Let Y X2 + 1. Compute E[Y] and sigma 2 y if

Explanation / Answer

P(Y>t) = P(X1>t,X2>t,X3>t) = (because of independence) = (1-t)^3 And here is another fact that is seldom used: E[Y] = integral {(1-t)^3 dt : t=0 to 1} = 1/4 For some value y, the probability of Y>y is: P(Y>y) = P(X1>y and X2>y and X3>y) = (1-y)^3 So P(y) = dP(Y>y)/dy = -3(1-y)^2 Now, EY = ? y P(y) dy / ? P(y) dy EY = ? -3 y*(1-y)^2 | [y=0 to 1] / ? -3(1-y)^2 | [y=0 to 1] = ? y^3 - 2y^2 + y dy | [y=0 to 1] / ? y^2 - 2y + 1 | [y=0 to 1] = 1/4 y^4 - 2/3 y^3 + 1/2 y^2 | [0 to 1] / (1/3 y^3 - y^2 + y | [0 to 1]) = (1/4 - 2/3 + 1/2) / (1/3 - 1 + 1) = 1/12 / 1/3 = 1/4 The answer is 1/4. (X>x)=(1-x), 0=y so P(Y>y)=(1-y)^3 giving F(y)=P(Y

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