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In a sequence of n Bernoulli trials, y successes were observed. The unknown probability of success π is estimated by π^=y/n a) Find lim n→[infinity]

​P( n( π^ −π)≤x), if π=0.5. a) Find lim n→[infinity] P( n (logit( π^ )−logit(π))≤t), if π=0.5.

1 Answer

3 votes

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.

User MrMythical
by
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