## Example Problems on Properties of a Random Sample

ST701 Homework 6 on Properties of a Random Sample

Problems: 4.54, 5.6, 5.22, 5.24, 5.42, 5.44, 7, 8

# 4.54

Find the pdf of $\prod_{i=1}^{n} X_i$ where the $X_i$’s are independent uniform(0, 1) random variables. (Hint: Try to calculate the cdf, and remember the relationship between uniforms and exponentials.)

Define $F_Z = P(\prod_{i=1}^{n} X_i < z)$. Then take the negative log of both sides of the inequality since the negative log of a uniform(0, 1) is an exp(1).

$F_Z = P(-\log(\prod_{i=1}^{n} X_i) \geq -\log(z)) = P( - \sum \log(X_i) \geq -\log(z)) = 1 - P( - \sum \log(X_i) < -\log(z))$

Since the $X_i$’s are iid exp(1), we can take a product of their MGFs to find the distribution.

\begin{align} M(t) & = \prod_{i = 1}^{n} M_{X_i}(t) \\ & = \prod_{i = 1}^{n} (\frac{ 1 }{ 1-\beta t })^\alpha \\ & = \prod_{i = 1}^{n} (\frac{ 1 }{ 1-1 \cdot t })^1 \\ & = (\frac{ 1 }{ 1- t })^n \end{align}

This is the MGF of a $Gamma(\alpha = n, \beta = 1)$.

\begin{align} F_Z & = 1 - P(Z \leq -\log(z)) \\ & = 1 - \int_{z=0}^{-\log(z)} \frac{ 1 }{ \Gamma(n) } x^n e^{-x/1} dx\\ & = 1 - \frac{ 1 }{ \Gamma(n) } \\ \\ f_Z & = -f_z(-\log(z)) \cdot \frac{ d }{ dz }(-\log(z)) \\ & = -\frac{ -1 }{ z }\frac{ 1 }{ \Gamma(n) } (-\log(z))^{n-1} e^{-(-\log(z))/1} \\ & = \frac{ 1 }{ z } \frac{ 1 }{ \Gamma(n) } (-\log(z))^{n-1} \cdot z \\ & = \frac{ 1 }{ \Gamma(n) } (-\log(z))^{n-1} & 0 < z < 1 \end{align}

# 5.6

If $X$ has pdf $f_X(x)$ and $Y$, independent of $X$, has pdf $f_Y(y)$, establish formulas, similar to (5.2.3), for the random variable $Z$ in each of the following situations.

## a.

$$Z = X - Y$$

Take $W = X$. Then, $Z = W - Y \Rightarrow Y = Z + W$.

$det \begin{bmatrix} \partial W & \partial Z \\ 1 & 0 \\ 1 & -1 \end{bmatrix} = -1 - 0$

Then,

$f_Z(z) = \int^{\infty}_{-\infty} f_X(w)\cdot f_Y(w - z) \cdot | -1 | dw$

## b.

$$Z = XY$$

Take $W = X$. Then, $Z = WY \Rightarrow Y = Z / W$.

$det \begin{bmatrix} \partial W & \partial Z \\ 1 & 0 \\ -z/w & 1/w \end{bmatrix} = 1/w$

Then,

$f_Z(z) = \int^{\infty}_{-\infty} f_X(w)\cdot f_Y(z/w) \cdot |1/w | dw$

## c.

$$Z = X/Y$$

Take $W = X$. Then, $Z = W / Y \Rightarrow Y = W / Z$.

$det \begin{bmatrix} \partial W & \partial Z \\ 1 & 0 \\ 1/z & -w/z^2 \end{bmatrix} = -w/z^2$

Then,

$f_Z(z) = \int^{\infty}_{-\infty} f_X(w)\cdot f_Y(w / z) \cdot | -w/z^2 | dw$

# 5.22

Let $X$ and $Y$ be iid n(0, 1) random variables, and define $Z = \min(X,Y)$. Prove that $Z^2 \sim \chi^2_1$.

Take $U_1 = X$ and $U_2 = Y$; then $Z = U_{(1)}$. Recall,

$f_{X_{(j)}}(x) = \frac{ n! }{ (j-1)!(n-j)! } f_X(x)[F_X(x)]^{j-1}[1-F_X(x)]^{n-j}.$ \begin{align} f_Z = f_{U_{(1)}} & = \frac{ 2! }{ (1-1)!(2-1)!} f_U(u) [F_U(u)]^{1-1} [1 - F_U(u)]^{2-1} \\ & = 2 f_U(u) [1-F_U(u)] \end{align}

Now take $W = Z^2$.

\begin{align} F_W(w) & = P(Z^2 \leq w) \\ & = P( -\sqrt{w} \leq Z \leq \sqrt{w})\\ & = P(Z \leq \sqrt{w}) - P(Z \leq - \sqrt{w}) \\ & = F_Z(\sqrt{w}) - F_Z(-\sqrt{w}) \\ \\ f_{Z^2} & = f_Z(\sqrt{w}) \frac{ 1 }{ 2 \sqrt{w} } - f_Z(-\sqrt{w}) \frac{ -1 }{ 2 \sqrt{w} } \end{align}

Notice that the normal PDF is symmetric so we know $F(X) = 1 - F(-X)$ and $f(x) = f(-x)$.

\begin{align} f_{Z^2} & = 2 \cdot f_U(\sqrt{ w }) [1-F_U(\sqrt{ w })] \frac{ 1 }{ 2 \sqrt{ w } } + 2 \cdot f_U(-\sqrt{w }) [1-F_U(-\sqrt{ w })] \frac{ 1 }{ 2 \sqrt{ w } } \\ & = f_U(\sqrt{ w }) \frac{ 1 }{ \sqrt{ w } } \Big( (1-F_U(\sqrt{ w })) + (1- F_U(- \sqrt{ w })) \Big) \\ & = f_U(\sqrt{ z }) \frac{ 1 }{ \sqrt{ w } } (1 - F_U(\sqrt{ w } + F_U(\sqrt{ w })) \\ & = f_U(\sqrt{ w }) \frac{ 1 }{ \sqrt{ w } } \\ & = \frac{ 1 }{ \sqrt{ w } } \frac{ 1 }{ \sqrt{ 2 \pi } } e^{-(\sqrt{ w })^2/2} \\ & = \frac{ 1 }{ \sqrt{ 2 \pi w } } e^{-w / 2} \end{align}

This is the PDF of a $\chi^2_1$.

# 5.24

Let $X_1, \dots ,X_n$ be a random sample from a population with pdf

$f_X(x) = \begin{cases} 1/\theta & \text{if } 0 < x < \theta \\ 0 & \text{otherwise.} \end{cases}$

Let $X_{(1)}< \dots < X_{(n)}$ be the order statistics. Show that $X_{(1)} / X_{(n)}$ and $X_{(n)}$ are independent random variables.

Take $U = \frac{ X_{1} }{ X_{(n)} }$ and $V = X_{(n)}$. Then $X_{(1)} = UV$.

$J = det \begin{bmatrix} \partial U & \partial V \\ V & U \\ 0 & 1 \end{bmatrix} = V$ \begin{align} f_{U,V} & = f_{X_{(1)}, X_{(n)}}(UV, V) \cdot | J | \\ & = |V| \cdot \frac{ n! }{ (1-1)! (n-1-1)! (n-n)! } f_X(UV) f_X(V) [F_X(U)]^{1-1} [F_X(V) - F_X(UV)]^{n-1-1} [1-F_X(V)]^{n-n} \\ & = V \frac{ n! }{ (n-2)! } f_X(UV) f_X(V)[F_X(V) - F_X(UV)]^{n-2} \\ & = V (n) (n-1) \frac{ 1 }{ \theta } \frac{ 1 }{ \theta } [\frac{ V-0 }{ \theta } - \frac{ UV - 0 }{ \theta }]^{n-2} \\ & = \frac{ n (n-1) }{ \theta^2} V \cdot V^{n-2} [\frac{ 1 }{ \theta } - \frac{ U }{ \theta }]^{n-2} \end{align}

Since this can be separated into the form $f_{U,V} = g(v) h(u)$, they are independent.

# 5.42

Similar to Example 5.5.11, let $X_1, X_2, \dots$ be iid and $X_{(n)} = \max_{1 \leq i \leq n} X_i$.

## a.

If $X_i \sim beta(1, \beta)$, find a value of $\nu$ so that $n^\nu (1 - X_{(n)})$ converges in distribution.

\begin{align} P(|X_{(n)} - 1 | \geq \epsilon) & = P(X_{(n)} \geq 1 + \epsilon) + P(X_{(n)} \leq 1 - \epsilon) \\ & = 0 + P(X_{(n)} \leq 1 - \epsilon) \\ & = P ( X_{i} \leq 1 - \epsilon, i = 1, \dots n) & iid \\ \end{align}

So we need to find the CDF of $X_i$.

\begin{align} F_{X_i} & = \int_{x=0}^{t} \frac{ \Gamma(\beta + 1) }{ \Gamma(\beta) \Gamma(1) } x^{1-1} (1-x)^{\beta - 1} dx \\ & = \int_{x=0}^{t} \frac{ \beta \Gamma(\beta ) }{ \Gamma(\beta) } (1-x)^{\beta - 1} dx \\ & = 1 - (1-t)^\beta \end{align} \begin{align} P(|X_{(n)} - 1 | \geq \epsilon) & = P ( X_{i} \leq 1 - \epsilon, i = 1, \dots n) \\ & = (1 - 1 - (1-\epsilon)^\beta )^n \\ & = (1 - \epsilon^\beta) ^n \end{align}

Take $\epsilon = \frac{ t^{1/\beta} }{ n^{1/\beta} }$.

\begin{align} P(|X_{(n)} - 1 | \geq \epsilon) & = P(X_{(n)} - \geq 1 - \frac{ t^{1/\beta} }{ n^{1/\beta} }) \\ & = P(n^{1/\beta}(1-X_{(n)} \leq t^{1/\beta}) \\ & \stackrel{d}{\rightarrow} 1 - e^{-t 1/\beta} \end{align}

Thus, $\nu = 1 / \beta$.

## b.

If $X_i \sim exponential(1)$, find a sequence $a_n$ so that $X_{(n)} - a_n$ converges in distribution.

\begin{align} \lim_{n \rightarrow \infty} F_{X_{(n)} - a_n} & = \lim_{n \rightarrow \infty} P(X_{(n)} - a_n \leq x) \\ & = \lim_{n \rightarrow \infty} P(X_{(n)} \leq x + a_n) \\ & = \lim_{n \rightarrow \infty} F_{X{(n)}}(x+a_n) \\ \\ F_{X_{(n)}} & = F_X(x)^n \\ & = \lim_{n \rightarrow \infty} (1 - e^{-(x + a_n)})^{n} \\ & = \lim_{n \rightarrow \infty} (1 - \frac{ n }{ n }e^{-(x + a_n)})^{n} \\ \end{align}

This converges to $e^{n e^{-(x+a_n)}}$ if $n e^{-(x + an)}$ converges, so choose $a_n$ accordingly, such as $a_n = \log(n)$.

# 5.44

Let $X_i, i = 1, 2, \dots$ be independent Bernoulli($p$) random variables and let $Y_n = \frac{ 1 }{ n } \sum_{i=1}^n X_i$.

## a.

Show that $\sqrt{n} (Y_n - p) \rightarrow n[0, p(1-p)]$ in distribution.

Since $E(X_i) = p$ and $Var(X_i) = p(p-1) > 0$ and $Y_n = \overline{ X }$, we can use the central limit theorem to say

$\sqrt{n} (Y_n - p) \rightarrow n[E(X_i) - p , p(1-p)] = N[0, p(1-p)].$

## b.

Show that for $p \neq 1/2$, the estimate of variance $Y_n(1-Y_n)$ satisfies $\sqrt{n} [Y_n ( 1 - Y_n)-p(1-p) ] \rightarrow n[0, (1-2p)^2 p (1-p)]$ in distribution.

Here we have $g(z) = z( 1 - z)$ so $g’(z) = 1 - 2z$. Thus,

$g'(p) = 1 - 2 (p).$

Since $p \neq 1/2$, we can use the first order delta method.

\begin{align} \sqrt{n} [Y_n ( 1 - Y_n)-p(1-p) ] & = \sqrt{ n }(g(Y_n) - g(p)) \\ & \stackrel{d}{\rightarrow} N(0, \sigma^2 [g'(p)]^2) \\ & = N(0, (p(1-p)) [1-2p]^2) \end{align}

## c.

Show that for $p = 1/2$, $n [Y_n(1-Y_n)-\frac{ 1 }{ 4 }] \rightarrow \frac{ 1 }{ 4 } \chi^2_1$ in distribution. (If this appears strange, note that $Y_n(1-Y_n) \leq 1/4$, so the left-hand is always negative. An equivalent form is $2n [\frac{ 1 }{ 4 } - Y_n(1-Y_n)] \rightarrow \chi^2_1$.)

Notice that plugging in $p = 1/2$ to our derivative above will give 0, so we should instead try the second order delta method. Notice that $g’‘(x) = 2$.

\begin{align} n [Y_n(1-Y_n)-\frac{ 1 }{ 4 }] & = n[g(Yn) - g(p)] \\ & \stackrel{d}{\rightarrow} \sigma^2 \frac{ g''(\theta) }{ 2 } \chi^{2}_{1} \\ & = 1/2 ( 1 - 1/2) 2/2 \chi^{2}_{1} \\ & = \frac{ 1 }{ 4 } \chi^{2}_{1} \end{align}

# 7

Let $X_1, X_2, \dots$ be a sequence of independent random variables with $E(X_i) = \mu$ and $V(X_i) = \sigma_i^2$. Show that if $n^{-2} \sum_{i=1}^n \sigma_i^2 \rightarrow 0$, then $\bar{X} \rightarrow \mu$ in probability.

We want to show $P( | \overline{X } - \mu | \geq \epsilon) = 0$. We can use Chebyshev’s inequality since $E(\overline{ X }) = \mu$.

\begin{align} P( | \overline{X } - \mu | \geq \epsilon) & \leq V(\overline{ X }) / \epsilon^2 \\ & = \frac{ 1/n^2 V(X_1 + \dots + X_n) }{ \epsilon^2 } \\ & = \frac{ 1/n^2 \sum \sigma_i^2 }{ \epsilon^2 } \end{align}

Since the numerator goes to 0,

$P( | \overline{ X } - \mu | \geq \epsilon ) = 0.$

So, $\overline{ X } \stackrel{p}{\rightarrow} \mu$.

# 8

Let $X_i$ be as in Problem 7 above but with $E(X_i) = \mu_i$ and $n^{-1} \sum_{i=1}^{n} \mu_i \rightarrow \mu$. Show that $\bar{X} \rightarrow \mu$ in probability.

\begin{align} P( | \overline{X } - \mu | \geq \epsilon) & = P( | \overline{X } + E(\overline{ X }) -E(\overline{ X })- \mu | \geq \epsilon) \\ & = P(| (\overline{ X } - E(\overline{ X })) + (E(\overline{ X }) - \mu) | \geq \epsilon) \end{align}

Take the expression in the absolute value to be $A_n$. Then take

\begin{align} B_n & = \{|\overline{ X } - E(\overline{ X })| \geq \frac{ \epsilon }{ 2 } \} \\ C_n & = \{| E(\overline{ X }) - \mu | \geq \frac{ \epsilon }{ 2 }\}. \end{align}

Notice

\begin{align} P(A_n) & \leq P(B_n \cup C_n) \\ & \leq P(B_n) + P(C_n) \\ \\ P(C_n) & = P(| \frac{ 1 }{ n }\sum (\mu_i) - \mu | \geq \frac{ \epsilon }{ 2 }) \\ & = 0 & \text{by convergence} \\ \\ P(B_n) & = P(|\overline{ X } - E(\overline{ X })| \geq \frac{ \epsilon }{ 2 }) \\ & \leq \frac{ V(\overline{ X }) }{ (\epsilon / 2)^2 } & \text{Chebyshev} \\ & = \frac{ 1/n^2 \sum \sigma_i^2 }{ \epsilon^2 / 2^2 } \\ & = 0. \end{align}

Thus,

$P(A_n) \rightarrow 0 \Rightarrow \overline{ X }_n \stackrel{p}{\rightarrow} \mu.$