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chugg_mt2-94_sol

chugg_mt2-94_sol - 1 EE464 Chugg Spring 1994 Midterm 2...

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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 non-negative 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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