The Gamma distribution Gamma(𝛼,𝛽) with paramters 𝛼>0 , and 𝛽>0 is defined by the density

Question

The Gamma distribution Gamma(𝛼,𝛽) with paramters 𝛼>0 , and 𝛽>0 is defined by the density

𝑓𝛼,𝛽(𝑥)=𝛽𝛼Γ(𝛼)𝑥𝛼−1𝑒−𝛽𝑥,for all𝑥≥0.
The Γ function is defined by

Γ(𝑠)=∫∞0𝑥𝑠−1𝑒−𝑥𝑑𝑥.
As usual, the constant 𝛽𝛼Γ(𝛼) is a normalization constant that gives ∫∞0𝑓𝛼,𝛽(𝑥)𝑑𝑥=1.

In this problem, let 𝑋1,…,𝑋𝑛 be i.i.d. Gamma variables with

𝛽=1𝛼for some 𝛼>0.
That is, 𝑋1,…,𝑋𝑛∼Gamma(𝛼,1𝛼) random variables for some 𝛼>0 . The pdf for 𝑋𝑖 is therefore

𝑓𝛼(𝑥)=1Γ(𝛼)𝛼𝛼𝑥𝛼−1𝑒−𝑥/𝛼,for all 𝑥≥0.

What is the limit, in probability, of the sample average 𝑋⎯⎯⎯⎯⎯𝑛 of the sample in terms of 𝛼 ?

𝑋⎯⎯⎯⎯⎯𝑛−→−−𝑛→∞𝐏 = ?

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Use the result from the previous problem to give a consistent estimator 𝛼̂ of 𝛼 in terms of 𝑋⎯⎯⎯⎯⎯𝑛 .

(Enter barX_n for 𝑋⎯⎯⎯⎯⎯𝑛 )

𝛼̂ =?

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For the Delta method to apply, at what value of 𝑥 does 𝑔 need to be continuously differentiable? (Your answer should be in terms of 𝛼 .)

𝑥=?
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What is its asymptotic variance of 𝛼̂ ?

𝑉(𝛼̂ )=

Answer 1

(a) alpha^2

(b) barX_n^(1/2)
(c1) alpha^2
(c2) Normal Distribution
(c3) ?
(d)?

Answer 2

I got green tick with alpha^2 for c1

Answer 3

Isolve=[ 1.89, 2.37]

Iplug-in=[ 1.88, 2.36]

Answer 4

Using the previous part, find confidence intervals for 𝛼 with asymptotic level 90% using both the “solving" and the “plug-in" methods. Use 𝑛=25 , and 𝑋⎯⎯⎯⎯⎯𝑛=4.5 .

(Enter your answers accurate to 2 decimal places. Use the Gaussian estimate 𝑞0.05≈1.6448 for best results.)

Isolve=[ , ]
Iplug-in=[ , ]

Answer 5

To find the limit in probability of the sample average 𝑋⎯⎯⎯⎯⎯𝑛, we can use the Law of Large Numbers. The Law of Large Numbers states that the sample average of a sequence of independent and identically distributed random variables converges in probability to the expected value of the random variable.

In this case, since 𝑋1, 𝑋2, ..., 𝑋𝑛 are i.i.d. Gamma random variables with parameters (𝛼, 1/𝛼), the expected value of each 𝑋𝑖 is 𝛼/𝛽 = 𝛼. Therefore, the limit in probability of 𝑋⎯⎯⎯⎯⎯𝑛 as 𝑛 approaches infinity is 𝐏 = 𝛼.

To obtain a consistent estimator 𝛼̂ of 𝛼 in terms of 𝑋⎯⎯⎯⎯⎯𝑛, we can simply use the sample average itself. Therefore, 𝛼̂ = 𝑋⎯⎯⎯⎯⎯𝑛.

For the Delta method to apply, the function 𝑔 needs to be continuously differentiable at a certain value 𝑥. In this case, we need to find the value of 𝑥 at which 𝑔 is continuously differentiable. However, the function 𝑔 is not provided in the given question, so we cannot determine the specific value of 𝑥.

Finally, to find the asymptotic variance of 𝛼̂, denoted as 𝑉(𝛼̂), we would need more information about the estimator 𝛼̂ and the distribution of the random variables 𝑋1, 𝑋2, ..., 𝑋𝑛. However, this information is not provided in the question, so we cannot determine 𝑉(𝛼̂).

Answer 6

The Gamma distribution Gamma(𝛼,𝛽) with paramters 𝛼>0 , and 𝛽>0 is defined by the density

(a) alpha^2

I got green tick with alpha^2 for c1

Isolve=[ 1.89, 2.37]

Using the previous part, find confidence intervals for 𝛼 with asymptotic level 90% using both the “solving" and the “plug-in" methods. Use 𝑛=25 , and 𝑋⎯⎯⎯⎯⎯𝑛=4.5 .

c3 and d?

c1 answer above is wrong