MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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Question

 

Figure attached below.

1. A bin of 5 transistors is known to contain 2 that are defective. The transistors are to be tested, one at a
time, until the defective ones are identified. Denote by X the number of tests made until the first defect
is identified and by Y the number of additional tests until the second defect is identified.
(a)  Find g(x|y) and draw a figure like Figure 4.3-1(b), depicting the conditional pmfs for
y = 1,2,3,and 4.
(b)  Find h(y|x) and draw a figure like Figure 4.3-1(c), depicting the conditional pmfs for
x = 1,2,3,and 4.
(c)  Find E [X|Y = 2] and Var (X|Y = 2)

146 Chapter 4 Bivariate Distributions For example, 4 1 P(X = 2|Y = 2) = g(2|2) = 12 3 Similarly, the conditional pmf of Y, given that X = x, is equal to x+ y h(y|x) = y = 1, 2, when x = 1, 2, or 3. 2x + 3 The joint pmf f(x, y) is depicted in Figure 4.3-1(a) along with the marginal pmfs. Now, if y = 2, we would expect the outcomes of x-namely, 1, 2, and 3–to occur in the ratio 3:4:5. This is precisely what g(x|y) does: 1+2 2+2 3+2 g(1|2) g(2|2) = 12 g(3|2) : 12 12 Figure 4.3-1(b) displays g(x|1) and g(x|2), while Figure 4.3-1(c) gives h(y|1), h(y|2), and h(y|3). Compare the probabilities in Figure 4.3-1(c) with those in Figure 4.3-1(a). They should agree with your intuition as well as with the formula for h(y|x). fy(y) y 12/21 2 • 3/21 • 4/21 • 5/21 9/21 1 • 2/21 • 3/21 • 4/21 1 2 3 5/21 7/21 9/21 fx(x) (a) Joint and marginal pmfs y y h(y|1) h(y|2) h(y|3) • 3/12 • 4/12 • 5/12 g(x|2) 2 • 3/5 • 4/7 • 5/9 1 • 2/9 • 3/9 • 4/9 g(x|1) 1 • 2/5 • 3/7 • 4/9 1 2 3 1 2 3 (b) Conditional pmfs of X, given y (c) Conditional pmfs of Y, given x Figure 4.3-1 Joint, marginal, and conditional pmfs Note that 0 <h(y|x). If we sum h(y|x) over y for that fixed x, we obtain Eh(y|x) = E f(x, y) fx(x) = 1. fx(x) y fx(x) y
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Transcribed Image Text:146 Chapter 4 Bivariate Distributions For example, 4 1 P(X = 2|Y = 2) = g(2|2) = 12 3 Similarly, the conditional pmf of Y, given that X = x, is equal to x+ y h(y|x) = y = 1, 2, when x = 1, 2, or 3. 2x + 3 The joint pmf f(x, y) is depicted in Figure 4.3-1(a) along with the marginal pmfs. Now, if y = 2, we would expect the outcomes of x-namely, 1, 2, and 3–to occur in the ratio 3:4:5. This is precisely what g(x|y) does: 1+2 2+2 3+2 g(1|2) g(2|2) = 12 g(3|2) : 12 12 Figure 4.3-1(b) displays g(x|1) and g(x|2), while Figure 4.3-1(c) gives h(y|1), h(y|2), and h(y|3). Compare the probabilities in Figure 4.3-1(c) with those in Figure 4.3-1(a). They should agree with your intuition as well as with the formula for h(y|x). fy(y) y 12/21 2 • 3/21 • 4/21 • 5/21 9/21 1 • 2/21 • 3/21 • 4/21 1 2 3 5/21 7/21 9/21 fx(x) (a) Joint and marginal pmfs y y h(y|1) h(y|2) h(y|3) • 3/12 • 4/12 • 5/12 g(x|2) 2 • 3/5 • 4/7 • 5/9 1 • 2/9 • 3/9 • 4/9 g(x|1) 1 • 2/5 • 3/7 • 4/9 1 2 3 1 2 3 (b) Conditional pmfs of X, given y (c) Conditional pmfs of Y, given x Figure 4.3-1 Joint, marginal, and conditional pmfs Note that 0 <h(y|x). If we sum h(y|x) over y for that fixed x, we obtain Eh(y|x) = E f(x, y) fx(x) = 1. fx(x) y fx(x) y
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