By Olive D.J.

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Similarly if VAR(Y |X = x) = v(x), then VAR(Y |X) = v(X) = E(Y 2|X) − [E(Y |X)]2. 8. Suppose that Y = weight and X = height of college students. Then E(Y |X = x) is a function of x. For example, the weight of 5 feet tall students is less than the weight of 6 feet tall students, on average. Notation: When computing E(h(Y )), the marginal pdf or pmf f(y) is used. When computing E[h(Y )|X = x], the conditional pdf or pmf f(y|x) is used. In a formula such as E[E(Y |X)] the inner expectation uses f(y|x) but the outer expectation uses f(x) since E(Y |X) is a function of X.

Find the support Y of Y . i) If t is an increasing function then, FY (y) = P (Y ≤ y) = P (t(X) ≤ y) = P (X ≤ t−1 (y)) = FX (t−1 (y)). ii) If t is a decreasing function then, FY (y) = P (Y ≤ y) = P (t(X) ≤ y) = P (X ≥ t−1 (y)) = 1 − P (X < t−1(y)) = 1 − P (X ≤ t−1 (y)) = 1 − FX (t−1 (x)). iii) The special case Y = X 2 is important. If the support of X is positive, use i). If the support of X is negative, use ii). If the support of X is (−a, a) (where a = ∞ is allowed), then FY (y) = P (Y ≤ y) = √ √ P (X 2 ≤ y) = P (− y ≤ X ≤ y) = √ y √ − y √ √ fX (x)dx = FX ( y) − FX (− y), 0 ≤ y < a2.

The support of continuous random variables Y1 and Y2 is the region where f(y1 , y2) > 0. The support is generally given by one to three inequalities such as 0 ≤ y1 ≤ 1, 0 ≤ y2 ≤ 1, and 0 ≤ y1 ≤ y2 ≤ 1. For each variable, set the inequalities to equalities to get boundary lines. For example 0 ≤ y1 ≤ y2 ≤ 1 yields 5 lines: y1 = 0, y1 = 1, y2 = 0, y2 = 1, and y2 = y1. Generally y2 is on the vertical axis and y1 is on the horizontal axis for pdf’s. To determine the limits of integration, examine the dummy variable used in the inner integral, say dy1 .