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Expectations Introduction to Probability & Statistics Expectations
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Expectations Mean : xpxxdiscrete x (), xfxdxxcontinuous(), EXxdFx[]()
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Example Consider the discrete uniform die example: x 1 2 3 4 5 6 p(x) 1 / 6 1 / 6 1 / 6 1 / 6 1 / 6 1 / 6 = E[X] = 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6) = 3.5
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[x] Expected Life For a producted governed by an exponential life distribution, the expected life of the product is given by Exedx x 0 ft )e x (x 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 00.511.522.53 Density X
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[x] Expected Life For a producted governed by an exponential life distribution, the expected life of the product is given by Exedx x 0 xedx x21 0 ft )e x (x 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 00.511.522.53 Density X
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[x] Expected Life For a producted governed by an exponential life distribution, the expected life of the product is given by Exedx x 0 xedx x21 0 ()2 2 1 ft )e x (x 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 00.511.522.53 Density X
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[x] Expected Life For a producted governed by an exponential life distribution, the expected life of the product is given by Exedx x 0 xedx x21 0 ()2 2 1 ft )e x (x 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 00.511.522.53 Density X 1/
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Variance 2 ()()xdFx 22 ()()xpx x 22 ()()xfxdx 22 Ex[()] =
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Example Consider the discrete uniform die example: x 1 2 3 4 5 6 p(x) 1 / 6 1 / 6 1 / 6 1 / 6 1 / 6 1 / 6 2 = E[(X- ) 2 ] = (1-3.5) 2 (1/6) + (2-3.5) 2 (1/6) + (3-3.5) 2 (1/6) + (4-3.5) 2 (1/6) + (5-3.5) 2 (1/6) + (6-3.5) 2 (1/6) = 2.92
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Property 22 ()()xfxdx ()()xxfx 22 2 xfx xfxdxfx 2 2 2()()()
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Property 22 ()()xfxdx ()()xxfx 22 2 xfx xfxdxfx 2 2 2()()() EXEX[][] 22 2
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Property 22 ()()xfxdx ()()xxfx 22 2 xfx xfxdxfx 2 2 2()()() EXEX[][] 22 2 EX[] 22
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Example Consider the discrete uniform die example: x 1 2 3 4 5 6 p(x) 1 / 6 1 / 6 1 / 6 1 / 6 1 / 6 1 / 6 2 = E[X 2 ] - 2 = 1 2 (1/6) + 2 2 (1/6) + 3 2 (1/6) + 4 2 (1/6) + 5 2 (1/6) + 6 2 (1/6) - 3.5 2 = 91/6 - 3.5 2 = 2.92
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Exponential Example For a producted governed by an exponential life distribution, the expected life of the product is given by ft )e x (x 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 00.511.522.53 Density X 222 1 xedx x () 222 EX[] 0.51 1/
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Exponential Example For a producted governed by an exponential life distribution, the expected life of the product is given by xedx x31 0 ft )e x (x 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 00.511.522.53 Density X 1/ 222 1 xedx x () 222 EX[] 1 2
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Exponential Example For a producted governed by an exponential life distribution, the expected life of the product is given by xedx x31 0 ()3 3 ft )e x (x 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 00.511.522.53 Density X 1/ 222 1 xedx x () 222 EX[] 1 2 1 2 = 1 2
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Properties of Expectations 1.E[c] = c 2.E[aX + b] = aE[X] + b 3. 2 (ax + b) = a 2 2 4.E[g(x)] = g(x)E[g(x)] X (x- ) 2 e -tx gxdFx()()
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Properties of Expectations 1.E[c] = c 2.E[aX + b] = aE[X] + b 3. 2 (ax + b) = a 2 2 4.E[g(x)] = g(x)E[g(x)] X (x- ) 2 2 e -tx (t) gxdFx()()
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Property Derviation Prove the property: E[ax+b] = aE[x] + b
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Property Derivation
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1 = a + b1 = aE[x] + b
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Class Problem Total monthly production costs for a casting foundry are given by TC = $100,000 + $50X where X is the number of castings made during a particular month. Past data indicates that X is a random variable which is governed by the normal distribution with mean 10,000 and variance 500. What is the distribution governing Total Cost?
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Class Problem Soln: TC = 100,000 + 50X is a linear transformation on a normal TC ~ Normal( TC, 2 TC )
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Class Problem Using property E[ax+b] = aE[x]+b TC = E[100,000 + 50X] = 100,000 + 50E[X] = 100,000 + 50(10,000) = 600,000
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Class Problem Using property 2 (ax+b) = a 2 2 (x) 2 TC = 2 (100,000 + 50X) = 50 2 2 (X) = 50 2 (500) = 1,250,000
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Class Problem TC = 100,000 + 50 X but, X ~ N(100,000, 500) TC ~ N(600,000, 1,250,000) ~ N(600000, 1118)
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