## Example Problems on Estimators

ST702 Homework 3 on Estimators

# 7.8

One observation, $X$, is taken from a $n(0, \sigma^2)$ population.

## a

Find an unbiased estimator of $\sigma^2$.

Notice that we can’t use the sample variance here because we only have one observation (and thus would be dividing by 0). Instead we will use $X^2$.

$E(X^2) = V(X) + E(X)^2 = \sigma^2 + \mu ^2 = \sigma^2 + 0 = \sigma^2$

## b

Find the MLE of $\sigma$.

\begin{align} L ( \sigma | \mathbf x) & = \prod_{i=1}^1 f(x_i | \sigma) \\ & = \frac{ 1 }{ \sqrt{ 2 \pi \sigma^2 } } \exp(\frac{ -x^2 }{ 2 \sigma^2 }) \\ l( \sigma | x) & = -\frac{ 1 }{ 2 } \log(2 \pi \sigma^2) - \frac{ x^2 }{ 2 \sigma ^2 } \\ \\ \frac{ \partial l }{ \partial \sigma } & = \frac{ -1 }{ \sigma} + \frac{ x^2 }{ \sigma^3 } \\ 0 & = \frac{ -1 }{ \sigma} + \frac{ x^2 }{ \sigma^3 } \\ \frac{ 1 }{ \sigma } & = \frac{ x^2 }{ \sigma^3 } \\ \sigma^2 & = x^2 \\ \sigma & = | x | \end{align}

Notice that

$\frac{ \partial ^2 l }{ \partial \sigma^2 } = \frac{ 2 }{ \sigma^2 } - \frac{ 3 x^2 }{ \sigma^4 }$

is decreasing at $\sigma = |x|$, so we found a maximum.

## c

Discuss how the method of moments estimator of $\sigma$ might be found.

\begin{align} E(X) & = 0 \\ E(X^2) & = \sigma^2 \\ x^2 & = \sigma^2 \\ \sigma & = | x | \end{align}

# 7.9

Let $X_1, \dots , X_n$ be iid with pdf

$f(x | \theta) = \frac{ 1 }{ \theta }, \ 0 \leq x \leq \theta, \ \theta > 0.$

Estimate $\theta$ using both method of moments and maximum likelihood. Calculate the means and variances of the two estimators. Which should be preferred and why?

Method of Moments

\begin{align} E(X) & = \frac{ 1 }{ 2 } (\theta - 0) \\ & = \frac{ \theta }{ 2 } \\ \hat{\theta} & = 2 \overline{ X } \\ \\ E(\hat{\theta}) & = E(2 \overline{ X }) \\ & = 2 E( \frac{ 1 }{ n } \sum X_i)\\ & = \frac{ 2 }{ n }(\frac{ 1 }{ 2 } \theta)^n \\ & = \theta \\ \\ V(\hat{\theta}) & = V(2 \overline{ X }) \\ & = 4 V(\frac{ 1 }{ n }\sum X_i) \\ & = \frac{ 4 }{ n^2 }(n \frac{ 1 }{ 12 } \theta ^2) \\ & = \frac{ \theta^2 }{ 3n } \end{align}

Maximum Likelihood

\begin{align} L(\theta | \mathbf X) & = \prod \frac{ 1 }{ \theta } I(0 \leq x \leq \theta) \\ &= \frac{ 1 }{ \theta^n } I(0 \leq x_{(1)}) I(x_{(n)} \leq \theta) \end{align}

Notice that this function decreases in $\theta$ and is maximized at $X_{(n)} = \theta$ (as is it zero for $\theta > X_{(n)}$). Recall the PDF of a maximum order statistic.

$f_{X_{(n)}}(x) = n f(x) F(x)^{n-1} = n \frac{ 1 }{ \theta } \Big( \frac{ x }{ \theta } \Big)^{n-1} = \frac{ n x^{n-1} }{ \theta^2 } I(0 \leq x \leq \theta)$ \begin{align} E(\hat{\theta}) & = \frac{ n }{ \theta^n } \int_{0}^{\theta} x \cdot x^{n-1} dx \\ &= \frac{ n }{ \theta^n } \frac{ \theta^{n+1} }{ n+1 } \\ & =\frac{ n \theta }{ n+1 } \\ \\ E(\hat{\theta}^2) & = \frac{ n }{ \theta^n } \int_{0}^{\theta} x^2 \cdot x^{n-1} dx \\ &= \frac{ n }{ \theta^n } \frac{ \theta^{n+2} }{ n+2 } \\ & =\frac{ n \theta^2}{ n+1 } \\ \\ V(\hat{\theta}) & = \frac{ n \theta^2}{ n+1 } - \frac{ n^2 \theta^2 }{ (n+1)^2 } \\ & = \frac{ n \theta^2 }{ (n+1)^2(n+2) } \end{align}

If we look at the variances, we notice that the MoM decreases at a rate of $\frac{ 1 }{ n }$ while the MLE decreases at a rate of $\frac{ 1 }{ n^2 }$. So, for large $n$ the MLE will offer a better estimate.

# 7.10

The independent random vaiables $X_1 , \dots , X_n$ have the common distribution

$P(X_i \leq x | \alpha, \beta) = \begin{cases} 0 & \text{if } x < 0 \\ (x/\beta)^\alpha & \text{if } 0 \leq x \leq \beta \\ 1 & \text{if } x > \beta, \end{cases}$

where the parameters $\alpha$ and $\beta$ are positive.

## a

Find the two-dimensional sufficient statistic for $(\alpha, \beta)$.

First we must differentiate the CDF to find the PDF.

$f( x | \alpha, \beta) = \begin{cases} 0 & \text{if } x < 0 \\ \frac{ \alpha x^{\alpha - 1} }{ \beta^\alpha } & \text{if } 0 \leq x \leq \beta \\ 0 & \text{if } x > \beta, \end{cases}$

Then we can proceed by the factorization theorem.

\begin{align} f(\mathbb x | \alpha, \beta) & = \prod_{i=1}^n \frac{ \alpha x_i^{\alpha-1} }{ \beta^\alpha } I(0 \leq x \leq \beta) \\ & = \frac{ \alpha^n }{ \beta^{\alpha n} } \prod(x_i^{\alpha - 1}) I(0 \leq x_{(1)}) I(x_{(n)} \leq \beta) \\ & = \frac{ \alpha^n }{ \beta^{\alpha n} } I(0 \leq x_{(1)}) \prod(x_i)^{\alpha - 1} I(x_{(n)} \leq \beta) \\ \end{align}

Thus our 2D sufficient statistic is $(\prod x_i, x_{(n)})$.

## b

Find the MLEs of $\alpha$ and $\beta$.

$MLE_\alpha$

\begin{align} L(\alpha | x , \beta) & = \prod \frac{ \alpha }{ \beta^\alpha } x_i^{\alpha -1} \\ & = \frac{ \alpha^n }{ \beta^{\alpha n} } \prod x_i^{\alpha -1} \\ \\ l(\alpha | x , \beta) & = \log \Big( \beta^{\alpha n} \prod x_i^{\alpha -1} \Big) \\ & = n \log(\alpha) + \alpha n \log(\beta) + (\alpha -1) \sum (\log(x_i)) \\ \\ \frac{ \partial l }{ \partial \alpha } & = \frac{ n }{ \alpha } + n \log(\beta) + \sum (\log(x_i)) \\ \\ 0 & = \frac{ n }{ \alpha } + n \log(\beta) + \sum (\log(x_i)) \\ \hat{\alpha} & = \frac{ n }{ n \log(\beta) - \sum (\log(x_i))} \end{align}

Notice that

$\frac{ \partial^2 l }{ \partial \alpha^2 } = \frac{ -n }{ \alpha^2 }$

is negative, so this likelihood is a maximum.

$MLE_\beta$

\begin{align} L(\beta | x , \alpha) & = \prod \frac{ \alpha }{ \beta^\alpha } x_i^{\alpha -1} \\ & = \frac{ \alpha^n }{ \beta^{\alpha n} } \prod x_i^{\alpha -1} I(0 \leq x_{(1)}) I(x_{(n)} \leq \beta) \end{align}

Notice that this decreases with $\beta$ as is zero if $\beta >x_{(n)}$, so it achieves its maximum at $x_{(n)}$.

## c

The length (in millimeters) of cuckoos’ eggs found in hedge sparrow nests can be modeled with the distribution. For the data

$22.0, 23.9, 20.9, 23.8, 25.0, 24.0, 21.7, 23.8, 22.8, 23.1, 23.1, 23.5, 23.0, 23.0,$

find the MLEs of $\alpha$ and $\beta$.

Since the MLE of $\beta$ is the max, it is $25$ for this data.

$MLE_\alpha = \frac{14}{14 \log (25)-\log (22*23.9*20.9*23.8*25*24*21.7*23.8*22.8*23.1*23.1*23.5*23.0*23.0)} = 12.5949$

# 7.19

Suppose that the random variables $Y_1, \dots , Y_n$ satisfy

$Y_i = \beta X_i + \varepsilon, \ i = 1, \dots , n$

where $x_1, \dots , x_n$ are fixed constants and $\varepsilon_1 , \dots , \varepsilon_n$ are iid $n(0, \sigma^2)$, $\sigma^2$ unknown.

## a

Find the two dimensional sufficient statistic for $(\beta, \sigma^2)$.

Notice

$\varepsilon_i = \underbrace{\beta x_i - Y_i}_{N(0, \sigma^2)}.$

We will proceed by the factorization theorem.

\begin{align} \prod \varepsilon_i & = \prod_{i=1}^n \frac{1}{\sqrt{2 \pi \sigma}} e^{-\frac{1}{2 \sigma^2}(\beta x_i - y_i)^2} \\ & = \frac{1}{(2 \pi \sigma)^{n/2}} e^{-\frac{1}{2 \sigma^2}\sum (\beta x_i - y_i)^2} \\ & = \frac{1}{(2 \pi \sigma)^{n/2}} e^{\frac{ -\beta \sum x_i^2 }{ 2 \sigma^2 } - \frac{ \sum y_i^2 }{ 2 \sigma^2 } + \frac{ 2 \beta \sum x_i y_i }{ 2 \sigma^2 }} \end{align}

Thus a sufficient statistic is $(\sum y_i^2, \sum x_i y_i)$.

## b

Find the MLE of $\beta$, and show that it is an unbiased estimator of $\beta$.

\begin{align} L(\beta | y) & = \prod_{i=1}^n \frac{1}{\sqrt{2 \pi \sigma}} e^{-\frac{1}{2 \sigma^2}(\beta x_i - y_i)^2} \\ & = \frac{1}{(2 \pi \sigma)^{n/2}} e^{-\frac{1}{2 \sigma^2}\sum (\beta x_i - y_i)^2} \\ \\ l( \beta | y ) & = \frac{ -n }{ 2 } \log(2 \pi \sigma^2) - \frac{ -\beta \sum x_i^2 }{ 2 \sigma^2 } - \frac{ \sum y_i^2 }{ 2 \sigma^2 } + \frac{ 2 \beta \sum x_i y_i }{ 2 \sigma^2 } \\ \\ \frac{ \partial l }{ \partial \beta } & = 0 -0 -0 - \frac{ 1 }{ \sigma^2 } \sum x_i y_i - \frac{ \beta }{ \sigma^2 } \sum x_i^2 \\ \\ 0 & = - \frac{ 1 }{ \sigma^2 } \sum x_i y_i - \frac{ \beta }{ \sigma^2 } \sum x_i^2 \\ \hat \beta & = \frac{ \sum x_i y_i }{ \sum x_i ^2} \end{align} \begin{align} E(\hat \beta) & = E\Big( \frac{ \sum x_i y_i }{ \sum x_i } \Big) \\ & = \frac{ \sum x_i E(y_i) }{ \sum x_i^2 } \\ & = \frac{ \sum x_i \cdot x_i \beta }{ \sum x_i^2 } \\ & = \beta \end{align}

## c

Find the distribution of the MLE of $\beta$.

We now know that $Y_i \sim N(\beta x_i, \sigma^2)$, so we know that $\sum c_i Y_i \sim N(\sum c_i \mu_i, \sum c_i^2 \sigma_i^2)$. So,

\begin{align} \hat \beta & \sim N\Big(\frac{ \sum x_i \cdot x_i \beta }{ \sum x_i^2 }, \frac{ \sum x_i^2 }{ \sum x_i^4)} \sigma^2\Big) \\ & \sim N( \beta, \frac{ \sigma^2 }{ \sum x_i^2 }) \end{align}

# 7.20

Consider $Y_1, \dots , Y_n$ as defined in Exercise 7.19.

## a

Show that $\sum Y_i / \sum x_i$ is an unbiased estimator of $\beta$.

$E\Big( \frac{ \sum Y_i }{ x_i } \Big) = \frac{ \sum E(Y_i) }{ \sum x_i } = \frac{ \sum x_i \cdot \beta}{ \sum x_i } = \beta$

## b

Calculate the exact variance of $\sum Y_i / \sum x_i$ and compare it to the variance of the MLE.

$Var \Big( \frac{ \sum Y_i }{ x_i } \Big) = \frac{ 1 }{ (\sum x_i)^2 } \sum V(Y_i) = \frac{ \sum \sigma_i^2 }{ (\sum x_i)^2} = \frac{ n \sigma^2 }{ n^2 \overline{ x}^2 } = \frac{ \sigma^2 }{ n \overline{ x }^2 }$

Since the numerators of the variances are the same, we can just look at the denominators. Notice that

$\sum x_i^2 - n \overline{ x }^2 = \sum x_i^2 - (\sum x_i)^2 = \sum (x_i - \overline{ x })^2 \geq 0.$

Thus, $\sum x_i^2 \geq n \overline{ x }^2$, or $V(\hat \beta) \leq Var \Big( \frac{ \sum Y_i }{ x_i } \Big)$.