Convergence in Probability am I doing something wrong?

• Artusartos
In summary, the conversation discusses the mean of a random sample from a distribution with the pdf f(x) = e^{-x}. The mgf of Y_n=\sqrt{n}(\bar{X_n}-1) is shown to be M_{Y_n}(t) = [e^{t/\sqrt{n}} - (t/\sqrt{n})e^{t/\sqrt{n}}]^{-n}, t < \sqrt{n}. The limiting distribution Y_n as n \rightarrow \infty is also found. However, there is a discrepancy in the answer due to the incorrect assumption that the distribution of the mean is Gamma(n,1) instead of 1/n*Gamma(n,1). This is because the sample mean is calculated
Artusartos

Homework Statement

Let $\bar{X_n}$ denote the mean of a random sample of size n from a distribution that has pdf $f(x) = e^{-x}$, $0<x<\infty$, zero elsewhere.

a) Show that the mgf of $Y_n=\sqrt{n}(\bar{X_n}-1)$ is $M_{Y_n}(t) = [e^{t/\sqrt{n}} - (t/\sqrt{n})e^{t/\sqrt{n}}]^{-n}$, $t < \sqrt{n}$ b) Find the limiting distribution $Y_n$ as $n \rightarrow \infty$

The Attempt at a Solution

a) We know that the pdf of $\bar{X_n}$ is $\Gamma(n,1)$

$M_{Y_n}(t) = (\frac{1}{1-t})^n$

$M_{Y_n}(t) = E(e^{tY_n}) = E(e^{t\sqrt{n}(\bar{X_n}-1)} = e^{-t\sqrt{n}}E(e^{t\sqrt{n}\bar{X_n}})$

MGF of $\bar{X_n}$ evaluated at $t\sqrt{n}$

$=e^{-t\sqrt{n}}(\frac{1}{1-t\sqrt{n}})^n = ((e^{t/\sqrt{n}})(1-t\sqrt{n}))^{-n}$

This does not match the answer they give because t is multiplied by $\sqrt{n}$ instead of being divided by it. But I just can't see where I went wrong...

The sum of n independent Exp(1) is Gamma(n,1), so if I'm not mistaken, the distribution of the mean is 1/n*Gamma(n,1) instead of Gamma(n,1).

csopi said:
The sum of n independent Exp(1) is Gamma(n,1), so if I'm not mistaken, the distribution of the mean is 1/n*Gamma(n,1) instead of Gamma(n,1).

Thanks a lot. Can you explain a little more about why it needs to be $\frac{1}{n}\Gamma(n,1)$?

Is it because the sample mean is $\frac{X_1 + X_2 + ... + X_n}{n}$? And since we know that the sum of n independent Exp(1) is Gamma(n,1), $X_1 + ... + X_n = \Gamma(n,1)$, right? Is that why we divide by n?

1. What is convergence in probability?

Convergence in probability is a concept in probability theory that describes the behavior of a sequence of random variables. It means that as the number of observations in the sequence increases, the probability that the values in the sequence will get closer to a certain value approaches one.

2. How is convergence in probability different from other types of convergence?

Convergence in probability is different from other types of convergence, such as pointwise convergence or almost sure convergence, because it only requires the values in the sequence to get closer to a certain value in probability, rather than in a specific way or with certainty.

3. Can convergence in probability be used to prove the convergence of a sequence?

Yes, convergence in probability can be used to prove the convergence of a sequence. If a sequence of random variables converges in probability, then it also converges in distribution, which is a stronger type of convergence.

4. What is the importance of convergence in probability in statistics?

Convergence in probability is important in statistics because it allows us to make inferences about the population based on a sample. It also helps us understand the behavior of estimators and their consistency as the sample size increases.

5. How can I determine if a sequence of random variables converges in probability?

There are several criteria for determining if a sequence of random variables converges in probability, such as the Law of Large Numbers and the Central Limit Theorem. These criteria involve checking the properties of the sequence, such as its mean and variance, to see if they meet certain conditions for convergence in probability.

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