Answer:
In both parts of the question, you're asked to find the limit as
n approaches infinity for certain probabilities involving the estimation of the unknown probability of success
π. Given that
=
0.5
π=0.5, we can simplify the expressions and apply limit properties.
Explanation:
Let's start with part (a):
a) Find
lim
�
→
∞
�
(
�
(
�
^
−
�
)
≤
�
)
lim
n→∞
P(n(
π
^
−π)≤x), if
�
=
0.5
π=0.5.
In a Bernoulli distribution, the variance of the estimator
�
^
π
^
is given by
Var
(
�
^
)
=
�
(
1
−
�
)
�
Var(
π
^
)=
n
π(1−π)
. Since
�
=
0.5
π=0.5, this variance simplifies to
Var
(
�
^
)
=
1
4
�
Var(
π
^
)=
4n
1
.
We can use the Central Limit Theorem (CLT) here. The CLT states that as
�
n approaches infinity, the distribution of the sample mean approaches a normal distribution with mean
�
μ (population mean) and variance
�
2
�
n
σ
2
, where
�
2
σ
2
is the population variance. Since we have
�
=
0.5
π=0.5 and
Var
(
�
^
)
=
1
4
�
Var(
π
^
)=
4n
1
, we can treat
�
^
π
^
as a sample mean of Bernoulli trials with
�
=
0.5
π=0.5.
Now, let's rewrite the expression
lim
�
→
∞
�
(
�
(
�
^
−
�
)
≤
�
)
lim
n→∞
P(n(
π
^
−π)≤x) as a z-score (standard score) and find the limit:
lim
�
→
∞
�
(
�
(
�
^
−
�
)
Var
(
�
^
)
≤
�
Var
(
�
^
)
)
lim
n→∞
P(
Var(
π
^
)
n(
π
^
−π)
≤
Var(
π
^
)
x
)
Substitute the values:
�
=
0.5
π=0.5 and
Var
(
�
^
)
=
1
4
�
Var(
π
^
)=
4n
1
:
lim
�
→
∞
�
(
2
�
(
�
^
−
0.5
)
1
4
�
≤
�
1
4
�
)
lim
n→∞
P(
4n
1
2n(
π
^
−0.5)
≤
4n
1
x
)
Simplify:
lim
�
→
∞
�
(
4
�
(
�
^
−
0.5
)
≤
2
�
)
lim
n→∞
P(4n(
π
^
−0.5)≤2x)
Notice that the left-hand side now resembles a z-score. As
�
n goes to infinity, the expression will converge to the standard normal distribution's cumulative distribution function (CDF). Therefore, the limit is:
lim
�
→
∞
�
(
4
�
(
�
^
−
0.5
)
≤
2
�
)
=
Φ
(
2
�
)
lim
n→∞
P(4n(
π
^
−0.5)≤2x)=Φ(2x)
where
Φ
Φ represents the standard normal cumulative distribution function.
This limit is not dependent on
�
π and will approach the value of
Φ
(
2
�
)
Φ(2x) as
�
n goes to infinity.
For part (b), the approach is similar, but it involves the logit transformation. The logit transformation of
�
^
π
^
is
logit
(
�
^
)
=
log
(
�
^
1
−
�
^
)
logit(
π
^
)=log(
1−
π
^
π
^
). You would follow a similar process of simplifying and finding the limit as
�
n approaches infinity.