MGF of standard normal dist~

2011-02-16 7:10 pm
If X is a random variable having a normal distribution μ with the mean and the standard deviation σ, show that the moment-generating function of Z=(X-μ)/σ is the moment-generating function of the standard normal distribution
更新1:

If the annual proportion of erroneous income tax returns filed with the IRS can be looked upon as a random variable having a beta distribution with α=2 and β=9, what is the probability that in any given year there will be fewer than 10% erroneous returns? The answer is 0.2643

回答 (1)

2011-02-16 10:10 pm
✔ 最佳答案
1) Moment generating function of X:

MX(t) = eμt+(σ^2t^2)/2

For Z = X/ σ - μ/σ, its MGF is:

MZ(t) = e-μt/σ MX(t/σ)

= e-μt/σ eμt/σ+(t^2)/2

= e(t^2)/2

which is an MGF of normal distribution with mean = 0 and standard deviation = 1

2) The Beta function is given by:

2∫(x = 0 → π/2) sin3 x cos17 x dx

= 2∫(x = π/2 → 0) sin2 x cos17 x d(cos x)

= 2∫(x = π/2 → 0) cos17 x - cos19 x d(cos x)

= 2[(cos18 x)/18 - (cos20 x)/20] (x = π/2 → 0)

= 1/90

So the pdf of the beta distribution is:

f(x) = 90x(1 - x)8

And the required probability is:

∫(x = 0 → 0.1) 90x(1 - x)8 dx

= 90∫(x = 0 → 0.1) (x - 1)(1 - x)8 + (1 - x)8 dx

= 90∫(x = 0 → 0.1) [(1 - x)8 - (1 - x)9] dx

= 90∫(x = 0.1 → 0) [(1 - x)8 - (1 - x)9] d(1 - x)

= 90 [(1 - x)9/9 - (1 - x)10/10] (x = 0.1 → 0)

= 0.2643
參考: 原創答案


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