(a) Show that the moment generating function is MX 1t 2 et t1 et 2 in .NET

Attach QR Code in .NET (a) Show that the moment generating function is MX 1t 2 et t1 et 2
(a) Show that the moment generating function is MX 1t 2 et t1 et 2
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(b) Find the mean and variance of X. S5-21. Let X1, X2, . . . , Xr be independent exponential random variables with parameter . (a) Find the moment generating function of Y X1 X2 p Xr. (b) What is the distribution of the random variable Y [Hint: Use the results of Exercise S5-20]. S5-22. Suppose that Xi has a normal distribution with mean 2 1, 2. Let X1 and X2 be independent. i and variance i,i (a) Find the moment generating function of Y X1 X2. (b) What is the distribution of the random variable Y S5-23. Show that the moment generating function of the chi-squared random variable with k degrees of freedom is MX (t) (1 2t) k 2. Show that the mean and variance of this random variable are k and 2k, respectively. S5-24. Continuation of Exercise S5-20. (a) Show that by expanding etX in a power series and taking expectations term by term we may write the moment generating function as MX 1t 2 E 1etX 2 1 r 1 t tr r! p 2
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(b) Use MX (t) to nd the mean and variance of X. S5-19. A random variable X has the exponential distribution f 1x2 e
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(a) Show that the moment generating function of X is MX 1t 2 a1 t b
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(b) Find the mean and variance of X. S5-20. A random variable X has the gamma distribution
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Thus, the coef cient of t r r! in this expansion is r , the rth origin moment. (b) Continuation of Exercise S5-20. Write the power series expansion for MX(t), the gamma random variable. (c) Continuation of Exercise S5-20. Find 1 and using the 2 results of parts (a) and (b). Does this approach give the same answers that you found for the mean and variance of the gamma random variable in Exercise S5-20
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CHEBYSHEV S INEQUALITY (CD ONLY)
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In 3 we showed that if X is a normal random variable with mean and standard deviation , P( 1.96 < X < 1.96 ) 0.95. This result relates the probability of a normal random variable to the magnitude of the standard deviation. An interesting, similar result that applies to any discrete or continuous random variable was developed by the mathematician Chebyshev in 1867.
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Chebyshev's Inequality
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For any random variable X with mean P10 X for c > 0. 0
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This result is interpreted as follows. The probability that a random variable differs from its mean by at least c standard deviations is less than or equal to 1 c2. Note that the rule is useful only for c > 1. For example, using c = 2 implies that the probability that any random variable differs from its mean by at least two standard deviations is no greater than 1 4. We know that for a normal random variable, this probability is less than 0.05. Also, using c 3 implies that the probability that any random variable differs from its mean by at least three standard deviations is no greater than 1 9. Chebyshev s inequality provides a relationship between the standard deviation and the dispersion of the probability distribution of any random variable. The proof is left as an exercise. Table S5-1 compares probabilities computed by Chebyshev s rule to probabilities computed for a normal random variable. EXAMPLE S5-8 The process of drilling holes in printed circuit boards produces diameters with a standard deviation of 0.01 millimeter. How many diameters must be measured so that the probability is at least 8 9 that the average of the measured diameters is within 0.005 of the process mean diameter Let X1, X2, . . . , Xn be the random variables that denote the diameters of n holes. The average measured diameter is X 1X1 X2 p Xn 2 n. Assume that the X s are independent random variables. From Equation 5-40, E1X 2 and V1X 2 0.012 n. Consequently, the 2 1 2 standard deviation of X is (0.01 n) . By applying Chebyshev s inequality to X , P1 0 X Let c = 3. Then, P10 X Therefore, P10 X 0 310.012 n2 1 2 2 8 9 0 310.012 n2 1 2 2 1 9 0 c10.012 n2 1 2 2 1 c2
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Table S5-1 Percentage of Distribution Greater than c Standard Deviations from the Mean c 1.5 2 3 4 Chebyshev s Rule for any Probability Distribution less than 44.4% less than 25.0% less than 11.1% less than 6.3% Normal Distribution 13.4% 4.6% 0.27% 0.01%
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Thus, the probability that X is within 3(0.012 n)1 2 of such that 3(0.012 n)1 2 0.005. That is, n EXERCISES FOR SECTION 5-10
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S5-25. The photoresist thickness in semiconductor manufacturing has a mean of 10 micrometers and a standard deviation of 1 micrometer. Bound the probability that the thickness is less than 6 or greater than 14 micrometers. S5-26. Suppose X has a continuous uniform distribution with range 0 x 10. Use Chebyshev s rule to bound the probability that X differs from its mean by more than two standard deviations and compare to the actual probability. S5-27. Suppose X has an exponential distribution with mean 20. Use Chebyshev s rule to bound the probability that X differs from its mean by more than two standard deviations and by more than three standard deviations and compare to the actual probabilities. S5-28. Suppose X has a Poisson distribution with mean 4. Use Chebyshev s rule to bound the probability that X differs from its mean by more than two standard deviations and by more than three standard deviations and compare to the actual probabilities. S5-29. Consider the process of drilling holes in printed circuits boards. Assume that the standard deviation of the diameters is 0.01 and that the diameters are independent. Suppose that the average of 500 diameters is used to estimate the process mean. (a) The probability is at least 15 16 that the measured average is within some bound of the process mean. What is the bound (b) If it is assumed that the diameters are normally distributed, determine the bound such that the probability is 15 16 that the measured average is closer to the process mean than the bound. S5-30. Prove Chebyshev s rule from the following steps. De ne the random variable Y as follows: Y (a) (b) (c) (d) e 1 0 if 0 X 0 otherwise c
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