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Probability: Problem Set 7/8 Solutions
Fall 2009 Instructor: W. D. Gillam
(1) Calculate the moment generating function gX (t) = E(etX^ ) when X has a gamma distribution with parameters α, β.
Solution. Make the change of variables y = (1 − βt)β−^1 x to compute:
gX (t) =
0
etxxα−^1 e−x/β βαΓ(α)
dx
= (1 − βt)−α
0
yα−^1 e−y Γ(α)
dy
= (1 − βt)−α.
(2) Use the above result to prove that when X 1 ,... , Xn are IIDRV with standard normal distribution, then X^21 + · · · + X n^2 has a chi squared distribution with n degrees of freedom. Do this by showing that the moment generating functions coincide. Recall that we proved this in class by another method (more-or-less explicit calculation of density functions...). (3) Calculate the moment generating function gX (t) when X has a normal distribution with expected value μ and variance σ^2. Use this to prove that a sum of independent random variables with (possibly different!) normal distributions again has a normal distribution. As in the previous problem, do this by equating moment generating functions instead of explicitly calculating the convolution of densities as we did in class.
Solution. Any normally distributed random variable is obtained from a random variable with standard normal distribution by a linear change of variables; keeping track of the effect of this change of variables on MGFs and using the computation for the standard normal that we did in class, we find
gX (t) = exp
μt +
σ^2 t^2
(4) A random variable X is said to have a t-distribution with n degrees of freedom if X has the same distribution as X 0 √ 1 n (X
2 1 +^ · · ·^ +^ X n^2 )
n
X 0
X 12 + · · · + X n^2
where X 0 ,... , Xn are IIDRV with standard normal distribution. This just differs by a factor of
n from Student’s distribution, but it is sometimes more natural. Calculate the variance of X and observe that it is > 1 (assuming that n > 2 so that it exists). Plot the standard normal density and the t-densities with n = 1, 2 , 3 on the same axes. Observe that the t-density is similar to the standard normal bell curve, but slightly more spread out, accounting for its slightly larger variance. 1
Figure 1. Graphs of the standard normal density (blue), and the t- densities with n = 1, 2 , 3 (red, yellow, green, respectively).
Solution. Since the variance of Student’s distribution is 1/(n − 2) (when n > 2), the variance of the t-distribution will be n/(n − 2) > 1 (for n > 2). The density fn(x) for the t-distribution with n degrees of freedom is
fn(x) =
πn
Γ( n+1 2 ) Γ( n 2 )
(1 + x^2 /n)
n+1 2.
Plots are shown in Figure 1. These were constructed using the command graph = Plot[{(1/Sqrt[2Pi])Exp[-x^2/2], (1/Sqrt[Pi1])(Gamma[1]/Gamma[1/2])(1 + x^2)^(-1), (1/Sqrt[Pi2])(Gamma[3/2]/Gamma[2/2])(1 + x^2/2)^(-3/2), (1/Sqrt[Pi3])(Gamma[2]/Gamma[3/2])*(1 + x^2/3)^(-2)}, {x, -3, 3}] in Mathematica. (5) When we sketched the proof of Stirling’s Formula in class, recall that we approxi- mated n! by writing
n! =
( (^) n
e
)n √ 2 πn + ǫ 1 + ǫ 2 ,
where the “errors” ǫi were supposed to be small (compared to the leading term) in the sense that
lim n→∞
( (^) n
e
)−n (^) ǫ √i 2 πn
for i = 1, 2. The error ǫ 1 arose when we approximated the integral
n! =
0
xne−xdx
Studies of the effects of copper on trout show the variance of log LC50 measure- ments to be around .4 with concentrations measured in mg / L. If n = 10 studies on LC50 for copper are to be carried out, find the probability that the sample mean of log LC50 will differ from the true population mean by no more than .5. If the EPA wants to be 95 percent sure that the sample mean of log LC50 will differ from the population mean by no more than .5, how many tests should they carry out? (10) Suppose X 1 ,... , Xn, Y 1 ,... , Ym are independent random variables, and the Xi are normally distributed with expected value μ 1 , variance σ^21 , while the Yi are normally distributed with expected value μ 2 and variance σ 22. Calculate the distribution of Z = X 1 + · · · + Xn + Y 1 + · · · + Ym. (11) Suppose the effects of copper on bass show the variance of the log LC50 measure- ments to be .8. If the population means of the log LC50 are the same for bass and trout, find the probability that, with random samples of ten log LC50 mea- surements for each species, the sample mean for trout exceeds the sample mean for bass by at least 1.
