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Unformatted text preview: 1 EE464  Chugg, Spring 1994 Midterm 2 Solution 1 Short Problems (50 points) This problem contains 6 unrelated short problems. (1a) The height of the randomly selected student may be modeled as a random variable with mean m X = 65 (in.). This random variable is clearly nonnegative so that Markov’s inequality can be used. Note that, since only the mean is known at this point, Markov’s bound is the only real option ⇒ Pr { X ( u ) ≥ 74 } ≤ 65 74 = 0 . 88 . Since the standard deviation is given as σ X = 4, this is just the “3 σ rule of thumb” ⇒ a = m X − 3 σ X = 65 − 3(4) = 53 b = m X + 3 σ X = 65 + 3(4) = 77 . The last part of this question is a twist on the Chebychev inequality. The trick is to manipulate the event to obtain the form required for Chebychev’s bound ⇒ Pr { X ( u ) ∈ [60 , 72] } ≥ Pr { X ( u ) ∈ [60 , 70] } = 1 − Pr { X ( u ) negationslash∈ [60 , 70] } = 1 − Pr { X ( u ) − 65  > 5 } ≥ 1 − parenleftbigg σ X 5 parenrightbigg 2 = 1 − 16 25 = 9 25 = 0 . 36 . (1b) The moment generating function for mean zero Gaussian is G X ( u ) ( s ) = E braceleftBig e sX ( u ) bracerightBig = exp parenleftBigg σ 2 X s 2 2 parenrightBigg . It follows that E braceleftBig e 3 X ( u ) bracerightBig = E braceleftBig e − 3 X ( u ) bracerightBig = exp parenleftBigg 9 σ 2 X 2 parenrightBigg . The expected value of the hyperbolic cosine is found by recalling that the expectation operator is linear ⇒ E { cosh(3 X ( u )) } = E braceleftBig 1 2 e 3 X ( u ) + 1 2 e − 3 X ( u ) bracerightBig = 1 2 E braceleftBig e 3 X ( u ) bracerightBig + 1 2 E braceleftBig e − 3 X ( u ) bracerightBig = exp parenleftBigg 9 σ 2 X 2 parenrightBigg . Also, a close review of the derivation given in class will convince you that the expec tation of e sX ( u ) converges for all real values of s , so that by plugging in s = 3 we have made no conceptual error. 2 K.M. Chugg  April 8, 1994 (1c) Y ( u ) is a Rayleigh random variable, which we have have not dealt with in great detail. The mean of Y ( u ) is m Y = integraldisplay ∞ yf Y ( u ) ( y ) dy = integraldisplay ∞ y 2 σ 2 exp parenleftBigg − y 2 2 σ 2 parenrightBigg dy. This integral is difficult to evaluate, however you should recognize a relation to the second moment of a Gaussian random variable ⇒ m Y = integraldisplay ∞ y 2 σ 2 exp parenleftBigg − y 2 2 σ 2 parenrightBigg dy = √ 2 π σ integraldisplay ∞ y 2 √ 2 πσ exp parenleftBigg − y 2 2 σ 2 parenrightBigg dy = parenleftBigg √ 2 π σ parenrightBigg parenleftbigg...
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 Spring '06
 Caire
 Variance, Probability theory, K.M. Chugg, dy exp

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