# Distribution Convergence

Let’s do a problem from Chapter 5 of All of Statistics.

Suppose $X_1, \dots X_n \sim \text{Uniform(0,1)}$. Let $Y_n = \bar{X_n}^2$. Find the limiting distribution of $Y_n$.

Note that we have $Y_n = \bar{X_n}\bar{X_n}$

Recall from Theorem 5.5(e) that if $X_n \rightsquigarrow X$ and $Y_n \rightsquigarrow c$ then $X_n Y_n \rightsquigarrow cX$.

So the question becomes does $X_n \rightsquigarrow c$ so that we can use this theorem? The answer is yes. Recall that from Theorem 5.4(b) $X_n \overset{P}{\longrightarrow} X$ implies that $X_n \rightsquigarrow X$. So if we can show that we converge to a constant in probability we know that we converge to the constant in distribution. Let’s show that $\bar{X}_n \overset{P}{\longrightarrow} c$. That’s easy. The law of large numbers tells us that the sample average converges in probability to the expectation. In other words $\bar{X}_n \overset{P}{\longrightarrow} \mathbb{E}[X]$. Since we are told that $X_i$ is i.i.d from a Uniform(0,1) we know the expectation is $\mathbb{E}[X] = .5$.

Putting it all together we have that:

$Y_n = \bar{X_n}^2$
$Y_n = \bar{X_n}\bar{X_n}$
$Y_n \rightsquigarrow \mathbb{E}[X]\mathbb{E}[X]$ (through the argument above)
$Y_n \rightsquigarrow (.5)(.5)$
$Y_n \rightsquigarrow .25$

We can also show this by simulation in R, which produces this chart:

Indeed we also get the answer 0.25. Here is the R code used to produce the chart above:

# Load plotting libraries
library(ggplot2)
library(ggthemes)

# Create Y = g(x_n)
g = function(n) {
return(mean(runif(n))^2)
}

# Define variables
n = 1:10000
Y = sapply(n, g)

# Plot
set.seed(10)
df = data.frame(n,Y)
ggplot(df, aes(n,Y)) +
geom_line(color='#3498DB') +
theme_fivethirtyeight() +
ggtitle('Distribution Convergence of Y as n Increases')