Solved – Variance of uniform distribution

uniform distributionvariance

I can see how one can compute the variance of a uniform distribution on $[a,b]$ using

$$ Var[X] = E[X^2] – E[X]^2 = \frac{(b-a)^2}{12}$$

as explained e.g. here: http://www.statlect.com/probability-distributions/uniform-distribution

However, how does one compute it using

$$ Var[X] = E[(X – E[X])^2] $$ ?

My approach would be transform $(X-E[X])^2$ to a new random variable $Y$ which follows a cdf $$G(y) = P(Y < y) = P((X-E[X])^2 < y) = P(X < \sqrt{y} + E[X]) = \frac{\sqrt{y} + E[X]-a}{b-a} $$

whenever $\sqrt{y} + E[X] \in [a,b]$, i.e., $y \in [a_Y, b_Y]$ where $a_Y = (\frac{3a-b}{2})^2$ and $b_Y = (\frac{b+a}{2})^2 $, and 0 otherwise.

Thus $g(y) = \frac{1}{2(b-a)\sqrt{y}}$ for $y \in [a_Y, b_Y]$ and 0 otherwise.

And then I would compute $$Var[X] = E[(X – E[X])^2] = E[Y] = \int_{a_Y}^{b_Y} y g(y) dy = \frac{1}{2(b-a)} \int_{a_Y}^{b_Y} \sqrt{y} dy$$

which does not yield the desired result.

What goes wrong with my transformation?

Best Answer

However, how does one compute it using $Var[X] = E[(X - E[X])^2]$?

You could just use the law of the unconscious statistician
$\operatorname{E}[g(X)] = \int_{-\infty}^\infty g(x) f_X(x) \, dx $ where $g$ is $(X-\mu)^2$

What goes wrong with my transformation?

Well, this is wrong, for one thing:

$P((X-E[X])^2 < y) = P(X < \sqrt{y} + E[X]) $

Related Solutions

Solved – Generating random samples from a custom distribution

It looks like you figured out that your code works, but @Aniko pointed out that you could improve its efficiency. Your biggest speed gain would probably come from pre-allocating memory for z so that you're not growing it inside a loop. Something like z <- rep(NA, nsamples) should do the trick. You may get a small speed gain from using vapply() (which specifies the returned variable type) instead of an explicit loop (there's a great SO question on the apply family).

> nsamples <- 1E5
> x <- runif(nsamples)
> f <- function(x, u) 1.5 * (x - (x^3) / 3) - u
> z <- c()
> 
> # original version
> system.time({
+ for (i in 1:nsamples) {
+   # find the root within (0,1) 
+   r <- uniroot(f, c(0,1), tol = 0.0001, u = x[i])$root
+   z <- c(z, r)
+ }
+ })
   user  system elapsed 
  49.88    0.00   50.54 
> 
> # original version with pre-allocation
> z.pre <- rep(NA, nsamples)
> system.time({
+ for (i in 1:nsamples) {
+   # find the root within (0,1) 
+   z.pre[i] <- uniroot(f, c(0,1), tol = 0.0001, u = x[i])$root
+   }
+ })
   user  system elapsed 
   7.55    0.01    7.78 
> 
> 
> 
> # my version with sapply
> my.uniroot <- function(x) uniroot(f, c(0, 1), tol = 0.0001, u = x)$root
> system.time({
+   r <- vapply(x, my.uniroot, numeric(1))
+ })
   user  system elapsed 
   6.61    0.02    6.74 
> 
> # same results
> head(z)
[1] 0.7803198 0.2860108 0.5153724 0.2479611 0.3451658 0.4682738
> head(z.pre)
[1] 0.7803198 0.2860108 0.5153724 0.2479611 0.3451658 0.4682738
> head(r)
[1] 0.7803198 0.2860108 0.5153724 0.2479611 0.3451658 0.4682738

And you don't need the ; at the end of each line (are you a MATLAB convert?).

Solved – Variance of Estimator (uniform distribution)

Why $\text{Cov}(X, Y) = 0$ if $X$ and $Y$ are independent?

By definition, $\text{Cov}(X, Y) = \mathbb{E}((X - \mu_X)(Y - \mu_Y))$. Hence, \begin{align*} \text{Cov}(X, Y) &= \mathbb{E}(XY - \mu_YX - \mu_XY + \mu_X \mu_Y) \\ &= \mathbb{E}(XY) - \mathbb{E}(\mu_YX) - \mathbb{E}(\mu_XY) + \mathbb{E}(\mu_X \mu_Y) ~ (\text{By linearity of expectation}) \\ &= \mathbb{E}(XY) - \mu_Y\mathbb{E}(X) - \mu_X\mathbb{E}(Y) + \mu_X \mu_Y ~ (\mu_X ~ \text{and} ~ \mu_Y ~ \text{are constants)} \\ &= \mathbb{E}(XY) - \mathbb{E}(X)\mathbb{E}(Y)! \\ \end{align*} Since $\mathbb{E}(XY) = \mathbb{E}(X)\mathbb{E}(Y)$ if $X$ and $Y$ are independent, $\text{Cov}(X, Y) = 0$.

Next, why $\text{Var}(\sum^n_{i = 1}X_i) = \sum^{n}_{i = 1}\text{Var}(X_i)$ if $X_1, X_2, \ldots, X_n$ are independent?

Take $n = 2$, \begin{align*} \text{Var}(X_1 + X_2) &= \mathbb{E}((X_1 + X_2)^2) - (\mathbb{E}(X_1 + X_2))^2 ~ (\text{definition}) \\ &=\mathbb{E}(X_1^2 + 2X_1X_2 + X_2^2) - (\mathbb{E}(X_1) + \mathbb{E}(X_2))^2 ~ (\text{expansion and linearity of expectation}) \\ &= \mathbb{E}(X^2_1) + 2\mathbb{E}(X_1X_2) + \mathbb{E}(X_2^2) - \mathbb{E}(X_1)^2 - 2 \mathbb{E}(X_1) \mathbb{E}(X_2) - \mathbb{E}(X_2)^2 \\ &= \mathbb{E}(X^2_1) - \mathbb{E}(X_1)^2 + \mathbb{E}(X^2_2) - \mathbb{E}(X_2)^2 ~ (\text{again, }\mathbb{E}(XY) = \mathbb{E}(X)\mathbb{E}(Y) \text{ by assumption}) \\ &= \text{Var}(X_1) + \text{Var}(X_2) \end{align*}

Back to the original question, what is $\text{Var}(T)$? \begin{align*} \text{Var}(2\frac{\sum^n_{i = 1}X_i}{n}) &= \frac{4}{n^2} \sum_{i = 1}^n \text{Var}(X_i) \\ &= \frac{4}{n^2} \times \frac{n\theta^2}{12} \\ &= \frac{\theta^2}{3n} \end{align*} Done!

Best Answer

Related Solutions

Solved – Generating random samples from a custom distribution

Solved – Variance of Estimator (uniform distribution)

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