Introduction to Probability and Statistics for Engineers and Scientists

Problems 277

maximizingf(x 1 ,...,xn|θ); that is, it is just the maximum likelihood estimate ofθ[when
it is constrained to be in (a,b)]. In other words, the maximum likelihood estimate equals
the mode of the posterior distribution when a uniform prior distribution is assumed. ■

If, rather than a point estimate, we desire an interval in whichθlies with a specified
probability — say 1−α— we can accomplish this by choosing valuesaandbsuch that

∫b

a

f(θ|x 1 ,...,xn)dθ= 1 −α

EXAMPLE 7.8d Suppose that if a signal of valuesis sent from location A, then the signal
value received at location B is normally distributed with meansand variance 60. Suppose
also that the value of a signal sent at location A is,a priori, known to be normally distributed
with mean 50 and variance 100. If the value received at location B is equal to 40, determine
an interval that will contain the actual value sent with probability .90.

SOLUTION It follows from Example 7.8b that the conditional distribution ofS, the signal
value sent, given that 40 is the value received, is normal with mean and variance given by

E[S|data]=

1/60

1/60+1/100

40 +

1/100

1/60+1/100

50 =43.75

Var(S|data)=

1

1/60+1/100

=37.5

Hence, given that the value received is 40, (S−43.75)/

√
37.5 has a unit standard
distribution and so

P

{

−1.645<

S−43.75

√

37.5

<1.645|data

}

=.90

or

P{43.75−1.645

√

37.5<S<43.75+1.645

√

37.5|data}=.95

That is, withprobability.90, the true signal sent lies within the interval (33.68, 53.82). ■

Problems..........................................................

1. LetX 1 ,...,Xnbe a sample from the distribution whose density function is

f(x)=

{

e−(x−θ) x≥θ

0 otherwise

Determine the maximum likelihood estimator ofθ.