This lecture shows how to derive confidence intervals for the variance of a normal distribution.
We analyze two different cases:
when the mean of the distribution is known;
when the mean is unknown.
For these two cases we derive the level of confidence and we show how to adjust it.
Two solved exercises can be found at the end of the lecture.
The theory needed to understand the derivations is presented in the page on interval estimation.
We start with the case in which the mean is known.
The sample is made of independent draws from a normal distribution.
Specifically, we observe the realizations of
independent random variables
,
...,
,
all having a normal distribution with:
known mean
;
unknown variance
.
We use the following estimator of
variance:
The confidence interval
iswhere
and
are strictly positive constants and
.
The choice of these constants is discussed below.
The coverage probability of the
confidence interval
is
where
is a Chi-square random variable with
degrees of freedom.
The coverage probability can be written
aswhere
we have
defined
In
the lecture on variance estimation, we
have shown that
has a Gamma distribution with parameters
and
,
given the assumptions on the sample made above. Multiplying a Gamma random
variable with parameters
and
by
,
we obtain a Chi-square random variable with
degrees of freedom, which in this case is the distribution of
.
The coverage probability does not depend on the unknown parameter
.
Therefore, the level of confidence
of the interval estimator
coincides with the coverage
probability:
where
is a Chi-square random variable with
degrees of freedom.
Note
thatwhere
is the distribution function of a Chi-square random variable with
degrees of freedom.
If
is the desired level of confidence, then we need to choose
and
so as to solve the
equation
As this is a single equation in two unknowns, there are infinitely many
choices of
and
that solve the equation.
Possible choices are:
set
,
which implies
and
;
numerically search for a couple of values
and
that not only solve the equation, but also minimize the length of the
confidence interval.
We now relax the assumption that the mean of the distribution is known.
The sample is made of the realizations of
independent variables
,
...,
,
all having a normal distribution with:
unknown mean
;
unknown variance
.
To construct interval estimators of the variance
,
we use the sample
mean
and
either the unadjusted
sample
variance
or
the adjusted sample
variance
We
consider the following confidence interval for the
variance:
where
and
are strictly positive constants and
.
Below we will see how to choose
and
.
The coverage probability
iswhere
is a Chi-square random variable with
degrees of freedom.
The coverage probability can be written
aswhere
we have
defined
In
the lecture on variance estimation, we
have proved that the unadjusted sample variance
has a Gamma distribution with parameters
and
.
Therefore,
has a Gamma distribution with parameters
and
where
But
a Gamma distribution with parameters
and
is the same as a Chi-square distribution with
degrees of freedom, which in this case is the distribution of
.
The coverage probability does not depend on the unknown parameters
and
.
Therefore, the level of confidence
is equal to the coverage
probability:where
is a Chi-square distribution with
degrees of freedom.
The level of confidence is the same found for the case of known mean. The only difference is in the number of degrees of freedom.
Therefore, the methods to choose
and
are those already illustrated above.
Below you can find some exercises with explained solutions.
Suppose that you observe a sample of 100 independent draws from a normal
distribution having known mean
and unknown variance
.
Denote the 100 draws by
,
...,
.
Suppose
that:
Find a confidence interval for
at 90% confidence.
Hint: the distribution
function
of a Chi-square random variable
with 100 degrees of freedom is such
that
Consider the confidence
intervalThe
level of confidence
is
where
is a Chi-square random variable with
degrees of freedom and
are strictly positive constants. If we
set
then
which
is equal to our desired level of confidence. Thus, the confidence interval for
is
Suppose that you observe a sample of 100 independent draws from a normal
distribution having unknown mean
and unknown variance
.
Denote the 100 draws by
,
...,
.
Suppose that their adjusted sample variance
is
Find a confidence interval for
.
Set the level of confidence at 99%.
Hint: a Chi-square random variable
with
degrees of freedom has a distribution function
such
that
Let the confidence interval
beThen,
the level of confidence
is
where
is a Chi-square random variable with
degrees of freedom and
are strictly positive constants. If we
set
then
which
is equal to the desired level of confidence. Thus, the confidence interval for
is
Please cite as:
Taboga, Marco (2021). "Confidence interval for the variance", Lectures on probability theory and mathematical statistics. Kindle Direct Publishing. Online appendix. https://www.statlect.com/fundamentals-of-statistics/set-estimation-variance.
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