Total: 50 pointsiid
~
Let X1, X2, … … …, Xn 2d Exp(0) be a sample from an exponential distribution with (unknown)
parameter 0 and pdf given by
fx(x; 0) = ½½e
1
-x/0
0
(Note that this is the “scale” parametrization for the Exponential distribution)
1. (2 points)
Derive the method of moments estimator MM. Is it biased or unbiased?
2. (5 points)
Derive the method of maximum likelihood estimator ÔMLE. Is it biased or unbiased?
3. (5 points)
Compute Var (MLE), the variance of the maximum likelihood estimator.
4. (8 points)
Using the fact that Exp(0) = F(1, 0), i.e. an Exponential distribution is just a Gamma distribution
with shape parameter 1, derive the exact sampling distribution for MLE.
5. (7 points)
Using your answer to part (4.), use the properties for transformations of random variables to find
the distribution of the new random variable
Y = H (ÔMLE, 0)
–
ÔMLE
Ꮎ
Does the distribution of Y depend on ?
6. (5 points)
Using your answer from part (5.) show that the exact two-sided confidence interval for with
confidence level (1 – a) is given by
nX
nX
CI(0; α) =
=
Пn,1;(1–a/2) ‘ În,1;α/2
where I’n,1,a denotes the a quantile of the Gamma distribution I (n, 1)Total: 50 points
iid
~
Let X1, X2, … … …, Xn 2d Exp(0) be a sample from an exponential distribution with (unknown)
parameter 0 and pdf given by
fx(x; 0) = ½½e
1
-x/0
0
(Note that this is the “scale” parametrization for the Exponential distribution)
1. (2 points)
Derive the method of moments estimator MM. Is it biased or unbiased?
2. (5 points)
Derive the method of maximum likelihood estimator ÔMLE. Is it biased or unbiased?
3. (5 points)
Compute Var (MLE), the variance of the maximum likelihood estimator.
4. (8 points)
Using the fact that Exp(0) = F(1, 0), i.e. an Exponential distribution is just a Gamma distribution
with shape parameter 1, derive the exact sampling distribution for MLE.
5. (7 points)
Using your answer to part (4.), use the properties for transformations of random variables to find
the distribution of the new random variable
Y = H (ÔMLE, 0)
–
ÔMLE
Ꮎ
Does the distribution of Y depend on ?
6. (5 points)
Using your answer from part (5.) show that the exact two-sided confidence interval for with
confidence level (1 – a) is given by
nX
nX
CI(0; α) =
=
Пn,1;(1–a/2) ‘ În,1;α/2
where I’n,1,a denotes the a quantile of the Gamma distribution I (n, 1)7. (5 points)
Using the fact that
2
г (k, 2) = x²k
for k € Z+, i.e., A Gamma distribution with parameters (k, 2) is a Chi-squared distribution with 2k
degrees of freedom, express the sampling distribution for MLE in terms of Xn
Show that the confidence interval in part (6) can be equivalently expressed as
2nX
2nX
CI(0; α)
”
2
X² 2n;1-a/2
X2n;a/2
where X²/2m,a
is the a quantile of the X2, distribution.
8. (10 points)
Using the central limit theorem for maximum likelihood estimators we know that the large sample
sampling distribution for MLE is approximately
ÔMLE ≈ N(0, Var(Ô MLE))
We also know that Var(0) = f(0), where f is some function of which you obtained in part 3.
Therefore,
0 – Ô MLE
√ƒ(0)
≈ N(0, 1).
A (1 – a) large sample confidence confidence interval using this approximation is given by
P(-30/2
0 – Ô MLE
√ƒ(0)
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