By Thabane L.
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Extra info for A closer look at the distribution of number needed to treat (NNT) a Bayesian approach (2003)(en)(6s)
Two chips are to be drawn successively, at random and without replacement. We want to compute the probability that the ﬁrst draw results in a red chip (C1 ) and that the second draw results in a blue chip (C2 ). It is reasonable to assign the following probabilities: P (C1 ) = 3 8 and P (C2 |C1 ) = 57 . 2679. 3. From an ordinary deck of playing cards, cards are to be drawn successively, at random and without replacement. The probability that the third spade appears on the sixth draw is computed as follows.
3) shows that P (X = 0) = 0 and we can assign fX (0) = 0 without changing the probabilities concerning X. The probability that a part has a lifetime between one and three years is given by 3 P (1 < X ≤ 3) = FX (3) − FX (1) = e−x dx. 1 That is, the probability can be found by FX (3) − FX (1) or evaluating the integral. 318. 1 shows that cdfs are right continuous and monotone. Such functions can be shown to have only a countable number of discontinuities. As the next theorem shows, the discontinuities of a cdf have mass; that is, if x is a point of discontinuity of FX , then we have P (X = x) > 0.
Suppose g(x) is strictly monotonically decreasing. 9) becomes FY (y) = 1 − FX (g −1 (y)). Hence, the pdf of Y is fY (y) = fX (g −1 (y))(−dx/dy). But since g is decreasing, dx/dy < 0 and, hence, −dx/dy = |dx/dy|. 8) is true in both cases. Henceforth, we refer to dx/dy = (d/dy)g −1 (y) as the Jacobian (denoted by J) of the transformation. In most mathematical areas, J = dx/dy is referred to as the Jacobian of the inverse transformation x = g −1 (y), but in this book it is called the Jacobian of the transformation, simply for convenience.