Consider a mixture distribution Z, where Fz(2) = W1Fx(2) + W2Fy(2). Var[x] = 5 Var[y] = 20 El[y] = 6 Var[y] = 20 Calculate the value of W that maximizes the variance of the mixture distribution, Var[z]. 0.50 0.55 0.60 0.65 0.70
Added by Rocio G.
Step 1
Now, we can substitute these values into the formula for Var[Z]: Var[Z] = W1^2 * 20 + (1 - W1)^2 * 20 + W1 * (1 - W1) * (5 - 6)^2 Var[Z] = 20 * W1^2 + 20 * (1 - 2W1 + W1^2) + W1 * (1 - W1) * 1 Var[Z] = 20 * W1^2 + 20 - 40W1 + 20 * W1^2 + W1 - W1^2 Var[Z] = 40 Show more…
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