Bar plot

Representing the sample with a single value

mean(pop_LD)

[1] 5.01

mean(pop_MD)

[1] 15.2

mean(pop_HD)

[1] 39.5

Variance: Quality of representation

var(pop_LD)

[1] 10.2

var(pop_MD)

[1] 79.6

var(pop_HD)

[1] 536

Standard Deviation

sd(pop_LD)

[1] 3.19

sd(pop_MD)

[1] 8.92

sd(pop_HD)

[1] 23.1

Standard Deviation is the “width” of population

We care about sd(x) because it tells us how close is the mean to most of the population

It can be proved that always \[\Pr(\vert x_i-\bar{\mathbf x}\vert\geq k\cdot\text{sd}(\mathbf x))\leq 1/k^2\]

In other words, the probability that “the distance between the mean \(\bar{\mathbf x}\) and any element \(x_i\) is bigger than \(k\cdot\text{sd}(\mathbf x)\)” is less than \((1/k^2)\)

This is Chebyshev’s inequality

It is always valid, for any probability distribution

(Later we will see better rules valid only sometimes)

It can also be written as \[\Pr(\vert x_i-\bar{\mathbf x}\vert\leq k\cdot\text{sd}(\mathbf x))\geq 1-1/k^2\]

The probability that “the distance between the mean \(\bar{\mathbf x}\) and any element \(x_i\) is less than \(k\cdot\text{sd}(\mathbf x)\)” is greater than \(1-1/k^2\)

Some examples of Chebyshev’s inequality

Another way to understand the meaning of this theorem is \[\Pr(\bar{\mathbf x} -k\cdot\text{sd}(\mathbf x)\leq x_i \leq \bar{\mathbf x} +k\cdot\text{sd}(\mathbf x))\geq 1-1/k^2\] Replacing \(k\) for some values, we get

\[\begin{aligned} \Pr(\bar{\mathbf x} -1\cdot\text{sd}(\mathbf x)\leq x_i \leq \bar{\mathbf x} +1\cdot\text{sd}(\mathbf x))&\geq 1-1/1^2 = 0\\ \Pr(\bar{\mathbf x} -2\cdot\text{sd}(\mathbf x)\leq x_i \leq \bar{\mathbf x} +2\cdot\text{sd}(\mathbf x))&\geq 1-1/2^2 = 0.75\\ \Pr(\bar{\mathbf x} -3\cdot\text{sd}(\mathbf x)\leq x_i \leq \bar{\mathbf x} +3\cdot\text{sd}(\mathbf x))&\geq 1-1/3^2 = 0.889 \end{aligned}\]

Sample average v/s size

Sample average v/s size

Log-log plot (for high density population)

plot(log(sd_sample_mean)~log(size))
model_HD <- lm(log(sd_sample_mean)~log(size))
lines(predict(model_HD)~log(size))

Linear model

coef(model_HD)

(Intercept)   log(size) 
      3.200      -0.516 

\[\log(\text{sd_sample_mean}) = 3.2 + -0.516\cdot\log(\text{size})\] \[\begin{aligned}\text{sd_sample_mean} & = \exp(3.2) \cdot\text{size}^{-0.516}\\ & = 24.529\cdot\text{size}^{-0.516} \end{aligned}\]

Models for all populations

\[\text{sd_sample_mean} = A\cdot \text{size}^B\]

Coefficient \(A\) is the standard deviation of the population
Coefficient \(B\) is -0.5

What does this mean

If we know the population standard deviation, we can predict the sample standard deviation

\[\text{sd(sample mean)} = \frac{\text{sd(population)}}{\sqrt{\text{sample size}}}\]

Law of Large Numbers

Using Chebyshev formula, we know that, with high probability \[\vert \text{mean(sample)} -\text{mean(population)}\vert < k\cdot\frac{\text{sd(population)}}{\sqrt{\text{sample size}}}\]

Therefore the population average is inside the interval \[\text{mean(sample)} \pm k\cdot\frac{\text{sd(population)}}{\sqrt{\text{sample size}}}\] (probably)

Some concepts

Random variable

number that depends on the outcome of a random process

Average of population

a fixed number, that we do not know and want to know

Average of a sample

a random variable, that we get from an experiment

Many things are averages

• concentration
• proportion
• density
• GC content
• frequency of events

What is the relationship between sample average and population average?

Can we learn the population average from the sample average?