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The multiplication rule for conditional PMFs. Let X, Y,
Chapter , Problem 36(choose chapter or problem)
The multiplication rule for conditional PMFs. Let X, Y, and Z be random variables. (a) Show that PX.y.z(x, y, z) = px(x)PYlx(y I x)PzIX.y(z I x, y). (b) How can we interpret this formula as a special case of the multiplication rule given in Section 1.3? (c) Generalize to the case of more than three random variables. Solution. (a) We have PX.y.Z(x, y, z) = P(X = x, Y = y, Z = z) = P(X = x)P(Y = y, Z = z I X = x) = P(X = x)P(Y = y I X = x)P(Z = z I X = x, Y = y) = px(x)PYlx(y I x)PZIX.y(z I x, y). (b) The formula can be written as P(X = x, Y = y, Z = z) = P(X = x)P(Y = y I X = x)P(Z = z I X = x, Y = y), which is a special case of the multiplication rule. (c) The generalization is PXI .... ,Xn (Xl, . . . ,Xn ) = PXI (xI )px2IxI (x2 1 xI ) . .. PXnIXI ..... Xn_1 (Xn I Xl , .. ,XnI ).
Questions & Answers
QUESTION:
The multiplication rule for conditional PMFs. Let X, Y, and Z be random variables. (a) Show that PX.y.z(x, y, z) = px(x)PYlx(y I x)PzIX.y(z I x, y). (b) How can we interpret this formula as a special case of the multiplication rule given in Section 1.3? (c) Generalize to the case of more than three random variables. Solution. (a) We have PX.y.Z(x, y, z) = P(X = x, Y = y, Z = z) = P(X = x)P(Y = y, Z = z I X = x) = P(X = x)P(Y = y I X = x)P(Z = z I X = x, Y = y) = px(x)PYlx(y I x)PZIX.y(z I x, y). (b) The formula can be written as P(X = x, Y = y, Z = z) = P(X = x)P(Y = y I X = x)P(Z = z I X = x, Y = y), which is a special case of the multiplication rule. (c) The generalization is PXI .... ,Xn (Xl, . . . ,Xn ) = PXI (xI )px2IxI (x2 1 xI ) . .. PXnIXI ..... Xn_1 (Xn I Xl , .. ,XnI ).
ANSWER:Step 1 of 4
Given that
X, Y, Z are the random variables.
a.