# Weak Convergence to Normal Distribution

• MHB
• joypav
In summary, the problem involves independent random variables with a specific probability distribution. The goal is to show that a certain expression converges weakly to a normal distribution as the number of variables increases. There is a discussion about the notation and a suggestion that it may relate to the Central Limit Theorem.
joypav
Problem:
Let $X_n$ be independent random variables such that $X_1 = 1$, and for $n \geq 2$,

$P(X_n=n)=n^{-2}$ and $P(X_n=1)=P(X_n=0)=\frac{1}{2}(1-n^{-2})$.

Show $(1/\sqrt{n})(\sum_{m=1}^{n}X_n-n/2)$ converges weakly to a normal distribution as $n \rightarrow \infty$.Thoughts:

My professor sent this problem over email and I am first off wondering about the notation. I think that the last line is meant to read $(\frac{1}{\sqrt{n}})(\frac{\sum_{m=1}^{n}X_m-n}{2})$? (I assume the summation index was simply a typo and that it is all over 2 not just the n?)

If that is the case, then I am thinking this is a consequence of the Central Limit Theorem.
Which has conclusion,

$\frac{S_n - \mu n}{\sigma \sqrt{n}} \implies \chi$

where $\chi$ is $N(0,1)$.

joypav said:
Problem:
Let $X_n$ be independent random variables such that $X_1 = 1$, and for $n \geq 2$,

$P(X_n=n)=n^{-2}$ and $P(X_n=1)=P(X_n=0)=\frac{1}{2}(1-n^{-2})$.

Show $(1/\sqrt{n})(\sum_{m=1}^{n}X_n-n/2)$ converges weakly to a normal distribution as $n \rightarrow \infty$.Thoughts:

My professor sent this problem over email and I am first off wondering about the notation. I think that the last line is meant to read $(\frac{1}{\sqrt{n}})(\frac{\sum_{m=1}^{n}X_m-n}{2})$? (I assume the summation index was simply a typo and that it is all over 2 not just the n?)

If that is the case, then I am thinking this is a consequence of the Central Limit Theorem.
Which has conclusion,

$\frac{S_n - \mu n}{\sigma \sqrt{n}} \implies \chi$

where $\chi$ is $N(0,1)$.
I am not a probabilist, but I think that your formula should read $\frac{1}{\sqrt{n}} \Bigl(\left(\sum_{m=1}^{n} X_m\right)- \frac{n}{2}\Bigr)$.

If $m$ is large, then $X_m$ takes each of the values $0$ and $1$ with a probability close to $\frac12$ (and the large value $m$ with a very small probability $m^{-2}$). So the mean value of $X_m$ will be close to $\frac12$, and the mean value of $S_n = \sum_{m=1}^{n} X_m$ will be close to $\frac n2$. This seems to make it plausible that your formula should somehow relate to the central limit theorem.

Last edited:

## 1. What is weak convergence to normal distribution?

Weak convergence to normal distribution is a concept in probability theory and statistics that describes the behavior of a sequence of random variables as they approach a normal distribution. It is also known as the central limit theorem, which states that the sum of a large number of independent and identically distributed random variables will tend towards a normal distribution.

## 2. How does weak convergence to normal distribution differ from strong convergence?

Weak convergence to normal distribution differs from strong convergence in that it only requires the convergence of the distribution of a sequence of random variables, rather than the convergence of the random variables themselves. Strong convergence, on the other hand, requires both the convergence of the distribution and the random variables.

## 3. What are the assumptions for weak convergence to normal distribution?

The assumptions for weak convergence to normal distribution include the independence of the random variables, identical distribution of the random variables, and finite variance. These assumptions are necessary for the central limit theorem to hold and for the convergence to a normal distribution to occur.

## 4. What are some applications of weak convergence to normal distribution?

Weak convergence to normal distribution has many applications in statistics and data analysis. It is often used in hypothesis testing, where the central limit theorem allows for the use of normal distribution-based tests even when the underlying distribution is not known. It is also commonly used in regression analysis and time series analysis.

## 5. How can one verify weak convergence to normal distribution?

There are several methods for verifying weak convergence to normal distribution, including graphical methods, such as the use of histograms or Q-Q plots, and statistical tests, such as the Kolmogorov-Smirnov test or the Shapiro-Wilk test. These methods can be used to compare the distribution of the sample data to a normal distribution and determine if the data is converging to a normal distribution.

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