Download Continuous Distributions

Chapter 5.6
From DeGroot & Schervish
Uniform Distribution
 f(x) = 1/ (b-a) for a≤ 𝑥 ≤ 𝑏
 For 0 ≤ 𝑥 ≤ 1, f(x) = 1
 m.g.f. of uniform distribution
1 𝑡𝑥
1 𝑡𝑥 1 1
tX
 ψ(t) = E(e )= 0 𝑒 .1 dx = 𝑒
= 𝑒𝑡 − 1
𝑡
0 𝑡
1
1 𝑡
𝑡
 ψ'(t) = − 2 𝑒 − 1 + . 𝑒
𝑡
𝑡
 ψ'(t) =

1
− 2
𝑡
1
2
1+𝑡
𝑡
3
= + +
𝑡2
6
𝑡2
+
2!
𝑡3
+
3!
1
𝑡
+ ⋯ − 1 + (1 + 𝑡 + ⋯ )
Mean and Variance
 ψ'(0) =


1
2
1
2𝑡
ψ''(t) = +
3
6
1
ψ''(0) =
3
 Var(X) = ψ''(0) – [ψ'(0)]2

1
3
1
4
= − =
1
12
The Normal Distributions
 The most widely used model for random variables with
continuous distributions is the family of normal
distributions.
 the random variables studied in various physical
experiments often have distributions that are
approximately normal.
 If a large random sample is taken from some
distribution, many important functions of the
observations in the sample will have distributions which
are approximately normal.
Properties of Normal Distributions
 A random variable X has the normal distribution with
mean μ and variance σ2 (−∞<μ<∞ and σ >0) if X has a
continuous distribution with the following p.d.f.:
The m.g.f. of Normal
Distribution
 Suppose Y = (X-)/, then X =  + Y. If the m.g.f. of Y,
ψY(t) = 𝑒
𝑡2
2
, we shall determine the m.g.f. of X
 ψX(t) = E(etX) = E(et( + Y))


= E(et + tY)) = et .E(e(t)Y)
= et . 𝑒
𝜎2 𝑡2
2
=𝑒
𝜎2 𝑡2
𝜇𝑡+ 2
Mean and Variance
 The mean and variance of the normal distribution are
μ and σ2, respectively.
 The first two derivatives of the m.g.f. of normal
distribution
The Shapes of Normal
Distributions
 The p.d.f. f (x|μ, σ2) of the normal distribution with
mean μ and variance σ2 is symmetric with respect to
the point x = μ.
 Therefore, μ is both the mean and the median of the
distribution.
 The p.d.f. f (x|μ, σ2) attains its maximum value at the
point x = μ.
The Shapes of Normal
Distributions
Linear Transformations
 If a random variable X has a normal distribution, then
every linear function of X will also have a normal
distribution.
 Theorem
 If X has the normal distribution with mean μ and
variance σ2 and if Y = aX + b, where a and b are given
constants and a ≠0, then Y has the normal distribution
with mean aμ + b and variance a2σ2.
Linear Transformations
 Proof
 If ψ denotes the m.g.f. of X, if ψY denotes the m.g.f. of
Y , then
 By comparing this expression for ψY with the m.g.f. of a
normal distribution given, we see that ψY is the m.g.f.
of the normal distribution with mean aμ + b and
variance a2σ2.
 Hence, Y must have this normal distribution.
The Standard Normal
Distribution
 The normal distribution with mean 0 and variance 1 is
called the standard normal distribution. The p.d.f. of
the standard normal distribution is usually denoted by
the symbol φ, and the c.d.f. is denoted by the symbol
.
Theorem
 For all x and all 0<p <1,
 Proof Since the p.d.f. of the standard normal distribution
is symmetric with respect to the point x = 0, it follows that
Pr(X ≤ x) = Pr(X≥−x) for every number x (−∞<x <∞).
 Pr(X ≤ x) = (x) and Pr(X ≥−x) = 1− (−x)
 Let x = −1(p) in the first equation and then apply the
function −1 to both sides of the equation.
Converting Normal
Distributions to Standard
 Let X have the normal distribution with mean μ and variance σ2. Let F
be the c.d.f. of X.
 Then Z = (X − μ)/σ has the standard normal distribution, and, for all x
and all 0<p <1,
 Proof Z = (X − μ)/σ has the standard normal distribution. Therefore,
 For second equation, let p = F(x) in the first equation and then solve for
x in the resulting equation.
Example
 Ortalama  = 220𝑉 olmak üzere günün değişik
zamanlarında voltaj ölçülüyor.  = 10V ve voltaj
ölçümlerinin normal dağılım olduğu biliniyorsa belirli
bir saatte voltajın 240’ın altına düşme olasılığı nedir?
Example
 Suppose that X has the normal distribution with mean 5 and standard
deviation 2. Determine the value of Pr(1<X <8).
 If we let Z = (X − 5)/2, then Z will have the standard normal
distribution and
 From the table at the end of this book, it is found that (1.5) = 0.9332
and (2) =0.9773. Therefore,
 Pr(1<X <8) = 0.9105.
Comparisons of Normal
Distributions
Linear Combinations of
Normally Distributed Variables
 Theorem
 If the random variables X1, . . . , Xk are independent
and if Xi has the normal distribution with mean μi and
variance σ2i (i = 1, . . . , k), then the sum X1 + . . . + Xk
has the normal distribution with mean μ1 + . . . + μk
and variance σ21+ . . . + σ2k.
Linear Combinations of
Normally Distributed Variables
 Proof
 Let ψi(t) denote the m.g.f. of Xi for i = 1, . . . , k, and let ψ(t) denote the
m.g.f. of X1 + . . . + Xk. Since the variables X1, . . . , Xk are independent,
then
 The m.g.f. ψ(t) can be identified as the m.g.f. of the normal
distribution for which the mean is 𝑘𝑖=1 𝜇𝑖 and the variance is
 Hence, X1 + . . . + Xk has the normal distribution.
𝑘
2
𝑖=1 𝜎𝑖
Corollary
Example
 Suppose that the heights, in inches, of the women in a
certain population follow the normal distribution with
mean 65 and standard deviation 1, and that the heights
of the men follow the normal distribution with mean
68 and standard deviation 3.
 Suppose also that one woman is selected at random
and, independently, one man is selected at random.
 Determine the probability that the woman will be
taller than the man.
Example
 Let W denote the height of the selected woman, and let M
denote the height of the selected man. Then the difference
W −M has the normal distribution with mean 65 − 68=−3
and variance 12 + 32 = 10. Therefore, if we let
 then Z has the standard normal distribution. It follows that
Corollary
Example
 Suppose that a random sample of size n is to be taken
from the normal distribution with mean μ and
variance 9.
 Determine the minimum value of n for which
Example
 the sample mean Xn will have the normal distribution
for which the mean is μ and the standard deviation is
3/n1/2. Therefore, if we let
 then Z will have the standard normal distribution. In
this example, n must be chosen so that
Example
 For each positive number x, it will be true that Pr(|Z| ≤ x) ≥
0.95 if and only if 1− (x) = Pr(Z > x) ≤ 0.025.
 From the table of the standard normal distribution at the
end the book, it is found that 1− (x) ≤ 0.025 if and only if
x ≥ 1.96.
 Therefore, the inequality will be satisfied if and only if
 Since the smallest permissible value of n is 34.6, the sample
size must be at least 35 in order that the specified relation
will be satisfied.
The Lognormal Distributions
 If log(X) has the normal distribution with mean μ and
variance σ2, we say that X has the lognormal
distribution with parameters μ and σ2.
m.g.f. of a Lognormal
Distribution
 The moments of a lognormal random variable are easy
to compute based on the m.g.f. of a normal
distribution.
 The definition of ψ is ψ(t) = E(etY ). Since Y = log(X),
we have
 It follows that E(Xt) = ψ(t) for all real t . In particular,
the mean and variance of X are