Homework 1

Solution.

1. This is a straightforward computation. Just keep in mind that to calculate $s^2$, you use $\frac{1}{n-1}$ instead of $\frac{1}{n}$.

   $$\boxed{ \begin{aligned} \mean{x} &\approx 7.2 \\ s^2 &\approx 4.2 \\ s &\approx 2.0 \end{aligned} }$$

2. There are $24$ observations (strictly) within one standard deviation from the mean, and $29$ (strictly) within two standard deviations from the mean. An efficient way to do this by hand is to calculate the range the observations need to lie in. For instance, for $x$ to be within one standard deviation from the mean, $x$ needs to satisfy

   $$\mean{x} - s \leq x \leq \mean{x} + s.$$

Python Code

data = [6.7, 5.4,  5.2, 6.0, 8.7, 6.0, 6.4,  8.3,  5.3, 5.9,
        7.6, 5.0,  6.9, 6.8, 4.9, 6.3, 5.0,  6.0,  7.2, 8.0,
        8.1, 7.2, 10.9, 9.2, 8.6, 6.2, 6.1, 14.1, 10.6, 8.4]

def sample_mean(data):
    n = len(data)
    return sum(data) / n


def sample_variance(data):
    x_bar = sample_mean(data)
    n = len(data)
    return sum([(x - x_bar) ** 2 for x in data]) / (n - 1)


def sample_standard_deviation(data):
    return sample_variance(data) ** 0.5


x_bar = sample_mean(data)
s2 = sample_variance(data)
s = sample_standard_deviation(data)

within_s = len([x for x in data if abs(x - x_bar) <= s])
within_2s = len([x for x in data if abs(x - x_bar) <= 2 * s])

print("x-bar:", x_bar)  # 7.233333333333333
print("s^2:  ", s2)     # 4.182298850574712
print("s:    ", s)      # 2.0450669550346543
print()
print("within  s of mean:", within_s)   # 24
print("within 2s of mean:", within_2s)  # 29

Solution.

For the mean,

For the sample variance,

Solution.

When sorted, the data is

Recall that to compute $\widetilde{\pi}_p$, we first decompose $\p{n+1}p = d + r$, where $d$ is an integer and $0 \leq r < 1$. In our case, $n = 30$, so

Thus, $\widetilde{\pi}_{0.25} = y_d + 0.75\p{y_{d+1} - y_d} = 5.9 + 0.75 \cdot \p{6.0 - 5.9} = 5.975$. The remaining percentiles are calculated similarly:

Suspected outliers are the ones that satisfy

and outliers satisfy

You can check by hand that $14.1$ is the only suspected outlier and that there are no outliers.

Python Code

import math


def sample_percentile(y, p):
    x = (len(data) + 1) * p
    d = math.floor(x)
    r = x - d

    # y[d] = y_{d+1} since arrays start at 0
    return y[d - 1] + r * (y[d] - y[d - 1])


y = sorted(data)

pi_25 = sample_percentile(y, 0.25)
pi_75 = sample_percentile(y, 0.75)
iqr = pi_75 - pi_25

pi_10 = sample_percentile(y, 0.1)
pi_90 = sample_percentile(y, 0.9)

suspected = [
    x
    for x in data
    if (pi_25 - 3 * iqr <= x and x <= pi_25 - 1.5 * iqr)
    or (pi_75 + 1.5 * iqr <= x and x <= pi_75 + 3 * iqr)
]

outliers = [x for x in data if x < pi_25 - 3 * iqr or pi_75 + 3 * iqr < x]

print("pi_25:", pi_25)  # 5.975
print("pi_75:", pi_75)  # 8.325000000000001
print("iqr:  ", iqr)    # 2.3500000000000014
print()
print("pi_10:", pi_10)  # 5.0200000000000005
print("pi_90:", pi_90)  # 10.460000000000003
print()
print("suspected outliers: ", suspected)  # [14.1]
print("outliers:           ", outliers)   # []

Solution.

First recall that being the $70^{\mathrm{th}}$ percentile means

1. We need at least $7$ of the $X_i$'s to be less than $27.3$, so

2. We need at least $5$ of the $X_i$'s to be less than $27.3$, and strictly fewer than $8$ of the $X_i$'s to be greater than $Y_8$, so

Solution.

1. Recall that the cdf of the order statistic $W_r$ is

   $$F_{W_r}\p{w} = \P\p{W_r \leq w} = \sum_{k=r}^n \binom{n}{k} F\p{w}^k \p{1 - F\p{w}}^{n-k} f\p{w},$$

   where $F, f$ are the cdf, pdf of the original random variables. In this case, $f\p{x} = 1$ for $0 < x < 1$ and $F\p{x} = x$ for $0 < x < 1$, so the cdf of $W_r$ is

   $$F_{W_r}\p{w} = \sum_{k=r}^n \binom{n}{k} w^k \p{1 - w}^{n-k}.$$

   The last step is a computation. There are two ways to do it: (i) by using the pdf given, or (ii) from the cdf alone. I'll do it via (ii) since I prefer using the cdf, and I think the computation will be a good review. (i) will be easier, however, and will just require step (1) below.

   I'll number the harder steps and explain them afterwards. If $f_{W_r}$ is the pdf, then

   $$\begin{aligned} \E\br{W_r^2} &= \int_0^1 w^2 f_{W_r}\p{w} \,\diff{w} \\ &= \left. w^2 F_{W_r}\p{w} \right\rvert_0^1 - 2 \int_0^1 wF_{W_r}\p{w} \,\diff{w} && \p{1} \\ &= 1 - 2 \int_0^1 \sum_{k=r}^n \binom{n}{k} w^{k+1} \p{1 - w}^{n-k} \,\diff{w} \\ &= 1 - 2 \sum_{k=r}^n \binom{n}{k} \int_0^1 w^{\p{k+2}-1} \p{1 - w}^{\p{n-k+1}-1} \,\diff{w} \\ &= 1 - 2 \sum_{k=r}^n \binom{n}{k} \frac{\Gamma\p{k + 2} \Gamma\p{n - k + 1}}{\Gamma\p{n + 3}} && \p{2} \\ &= 1 - 2 \sum_{k=r}^n \frac{n!}{k!\p{n - k}!} \frac{\p{k + 1}!\p{n - k}!}{\p{n + 2}!} \\ &= 1 - 2 \sum_{k=r}^n \frac{k + 1}{\p{n + 2}\p{n + 1}} \\ &= 1 - \frac{2}{\p{n + 1}\p{n + 2}} \sum_{k=r}^n \p{k + 1} && \p{3} \\ &= 1 - \frac{2}{\p{n + 1}\p{n + 2}} \frac{\p{n + r + 2}\p{n - r + 1}}{2} \\ &= \frac{r\p{r + 1}}{\p{n + 1}\p{n + 2}}. \end{aligned}$$

   (1) is integration by parts with

   $$\begin{cases} u = w^2 \\ \diff{v} = f_{W_r}\p{w} \,\diff{w} \\ \end{cases} \implies \begin{cases} \diff{u} = 2w \,\diff{w} \\ v = F_{W_r}\p{w}. \end{cases}$$

   (2) is from the pdf of a $\operatorname{Beta}\p{\alpha, \beta}$ distribution, which tells us

   $$\int_0^1 x^{\alpha-1} \p{1 - x}^{\beta-1} \,\diff{x} = \frac{\Gamma\p{\alpha} \Gamma\p{\beta}}{\Gamma\p{\alpha + \beta}},$$

   where $\Gamma\p{n} = \p{n-1}!$ for positive integers $n$.

   (3) is the sum of an arithmetic sequence. The result is a little different than what's there since the sum starts at $k = r$ instead of $k = 0$, but you can fix that by using

   $$\sum_{k=r}^n a_k = \sum_{k=0}^n a_k - \sum_{k=0}^{r-1} a_k.$$

2. Recall that for any random variable $X$,

   $$\Var X = \E\br{X^2} - \p{\E\br{X}}^2.$$

   We already know that $\E\br{W_r} = \frac{r}{n+1}$ and we computed the second moment in part 1, so plugging everything in,

   $$\Var W_r = \frac{r\p{r+1}}{\p{n+1}\p{n+2}} - \p{\frac{r}{\p{n+1}}}^2 = \boxed{\frac{r\p{n-r+1}}{\p{n+1}^2\p{n+2}}}.$$

   (Side note: The expectation and variance of $W_r$ agree with those of a $\operatorname{Beta}\p{r, n - r + 1}$ random variable!)