• When population is small, sampling will change the population
  • We do not do this in this course
• When population is big, sampling has a very small effect
  • The difference between 1E8 and 1E8-1 is not important
• When populations have infinite size, taking a sample has no effect
• When we replace the sample, there is no effect on the population
  • Sampling with replacement is the same as having an infinite population

• These simulations are called “Monte-Carlo Methods”
• This allow us to explore cases that have too many combinations
  • We cannot see all possible genes of length 1000bp
  • There are 4¹⁰⁰⁰ = 2²⁰⁰⁰ = 2^10x200 ≈ 10^3x200 = 10⁶⁰⁰ combinations
  • The age of the universe is ≈ 4.32x10¹⁷ seconds

Testing all possible cases is impossible

Random sampling allows us to get an idea of all possible cases

More simulations give better approximations, but take longer time

This is one of the most common uses of computers in Science

Find the population width for different values of \(k\)

• at least 75% of the population is near the average, by no more than 2 times the standard deviation \[\Pr(\vert x_i-\bar{\mathbf{x}}\vert\leq 2\cdot\text{sd}(\mathbf x))\geq 1-1/2^2\] \[\Pr(\bar{\mathbf{x}} -2\cdot\text{sd}(\mathbf x)\leq x_i \leq \bar{\mathbf{x}} +2\cdot\text{sd}(\mathbf x)) \geq 0.75\]
• at least 88.9% of the population is near the average, by less than 3 times the standard deviation \[\Pr(\bar{\mathbf{x}} -3\cdot\text{sd}(\mathbf x)\leq x_i \leq \bar{\mathbf{x}} +3\cdot\text{sd}(\mathbf x)) \geq 0.889\]

• You can take the average of a sample to estimate the population average
• The sample average is a random variable. Changes on every experiment
• When the sample is bigger, the sample average will be closer to the population average
  • This is called Law of Large Numbers
• Moreover, the sample average of a big sample will follow a Normal distribution
  • This is called Central Limit Theorem

≈95% of normal population is between \(-2\cdot\text{sd}(\mathbf x)\) and \(2\cdot\text{sd}(\mathbf x)\)
≈99% of normal population is between \(-3\cdot\text{sd}(\mathbf x)\) and \(3\cdot\text{sd}(\mathbf x)\)

• The Chebyshev rule is always valid, but pessimist
  • confidence intervals are big
• If the probability distribution is Normal, we can have better confidence intervals
• Not all probabilities are Normal
• When the random process is a sum of many parts, then we may have a Normal distribution
  • That happens with experimental measurements
  • Also happens in many biological processes

If we have 95% of population in the center, then we have 2.5% to the left and 2.5% to the right

We can find the \(k\) value using R

qnorm(0.025)

[1] -1.96

qnorm(0.975)

[1] 1.96

If we have \(1-\alpha\) of population in the center, then we have \(\alpha/2\) to the left and \(\alpha/2\) to the right

qnorm(alpha/2)
qnorm(1-alpha/2)

Now you can find any interval

You may have noticed that we never get 100% confidence

That is a fact of life. We have to accept

To have very high confidence, we need wide intervals

But wide intervals are less useful

Distribution: probability of each outcome

When the outcome is numeric (i.e. when it is a random variable), the distribution is

• a vector, for discrete outcomes
• a function, for continuous outcomes (real valued random variables)

We will represent population variance like this \[\mathbb V(X)=\langle(X-\langle X\rangle)^2\rangle\] Do not confuse it with sample variance \(\text{var}(x)\)

Same idea, but different context

It is good to know that \[\mathbb V(X)=\langle(X-\langle X\rangle)^2\rangle=\langle X^2\rangle-\langle X\rangle^2\]

Variance is the average of squares minus the square of average

Exercise: verify that this is true

If \(X\) and \(Y\) are two independent random variables, then \[\mathbb V(X+Y) = \mathbb V(X) + \mathbb V(Y) \] i.e. the variance of the sum is the sum of the variances

If \(k\) is a fixed (non-random) number, then \[\mathbb V(k\, X) = k^2\, \mathbb V(X)\] i.e. the variance is multiplied by the square

Exercise: verify that this is true

Variance is easy to calculate and has good properties. But the values are too big.

In practice we use the square root of variance, called standard deviation

We will represent population standard deviation like this \[\mathbb S(X)=\sqrt{\langle(X-\langle X\rangle)^2\rangle}=\sqrt{\langle X^2\rangle-\langle X\rangle^2}\] Do not confuse it with sample standard deviation \(\text{sd}(x)\)

If \(X\) and \(Y\) are two independent random variables, then \[\mathbb S(X+Y) = \sqrt{\mathbb{S}(X)^2 + \mathbb{S}(Y)^2} \] i.e. the same formula of Pythagoras’ theorem

If \(k\) is a fixed (non-random) number, then \[\mathbb S(k\, X) = k\, \mathbb S(X)\] i.e. the variance is multiplied by the square

Exercise: verify that this is true

Imagine that we repeat the experiment \(m\) times. In other words, the sample size is \(m\)

We get the values \(x_1, x_2,\ldots, x_m.\) All result from the same experiment, with the same probabilities

We want to see what happens with \[\sum_{i=1}^m x_i\]

Using the rules we discussed before, we have \[\left\langle\sum_{i=1}^m x_i\right\rangle=\sum_{i=1}^m \langle X\rangle=m\,\langle X\rangle\] The average of the sum is \(m\) times the average of each value.

We also have \[\mathbb V\left(\sum_{i=1}^m x_i\right)=\sum_{i=1}^m \mathbb V( X)=m\,\mathbb V(X)\] The variance of the sum is \(m\) times the variance of each value. Finally, \[\mathbb S\left(\sum_{i=1}^m x_i\right)=\sqrt{\mathbb V\left(\sum_{i=1}^m x_i\right)}=\sqrt{m}\,\mathbb S(X)\]

If now we take the average of each sample, we have \[\left\langle\frac{1}{m}\sum_{i=1}^m x_i\right\rangle=\sum_{i=1}^m \frac{1}{m}\langle X\rangle=\langle X\rangle\] The population average of the sample means is the population average of each value.

We also have \[\mathbb V\left(\frac{1}{m}\sum_{i=1}^m x_i\right)=\sum_{i=1}^m \frac{1}{m^2}\mathbb V( X)=\frac{1}{m}\mathbb V(X)\] The variance of the sample means is \(1/m\) times the variance of each value.

Finally, \[\mathbb S\left(\frac{1}{m}\sum_{i=1}^m x_i\right)=\sqrt{\frac{1}{m}\mathbb V\left(\sum_{i=1}^m x_i\right)}=\frac{\mathbb S(X)}{\sqrt{m}}\]

If we know the average and standard deviation of population, we can say that

• at least 75% of the population are close to the average, by less than 2 times the standard deviation \[\Pr\left(\vert \bar X_m-\langle X\rangle\vert\leq 2\frac{\mathbb S(X)}{\sqrt{m}}\right)\geq 1-1/2^2\] \[\Pr\left(\langle X\rangle -2\frac{\mathbb S(X)}{\sqrt{m}}\leq \bar X_m \leq \langle X\rangle +2\frac{\mathbb S(X)}{\sqrt{m}}\right) \geq 0.75\]

• At least 88.9% of the population are close to the average, by less than 3 times the standard deviation \[\Pr\left(\langle X\rangle -3\frac{\mathbb S(X)}{\sqrt{m}}\leq \bar X_m \leq \langle X\rangle +3\frac{\mathbb S(X)}{\sqrt{m}}\right) \geq 0.889\]

• 95% of the sample averages are close to the population average, at less than 2 times the standard deviation \[\Pr\left(\vert \bar X_m-\langle X\rangle\vert\leq 2\frac{\mathbb S(X)}{\sqrt{m}}\right)\approx 0.95\] \[\Pr\left(\bar X_m -2\frac{\mathbb S(X)}{\sqrt{m}}\leq \langle X\rangle\leq \bar X_m +2\frac{\mathbb S(X)}{\sqrt{m}}\right) \approx 0.95\]

• 99% of the sample averages are close to the population average, at less than 3 times the standard deviation \[\Pr\left(\bar X_m -3\frac{\mathbb S(X)}{\sqrt{m}}\leq \langle X\rangle \leq \bar X_m+3\frac{\mathbb S(X)}{\sqrt{m}}\right) \approx 0.99\]

All we said before is true, but cannot be used directly

Because we do not know the population variance

Thus, we also ignore the population standard deviation

What can we do instead?

The solution is to use the standard deviation of the sample

(do not confuse it with standard deviation of the sample means)

But we have to pay a price: lower confidence

Here we use the sample standard deviation to approximate the population standard deviation

As we have seen, if the sample is small, these two values may be different

Thus, the Student’s distribution depends on the sample size

The key idea is that the sample has \(m\) elements, but they are constrained by 1 value: the sample average

We say that we have \(m-1\) degrees of freedom

If we have 95% of population in the center, then we have 2.5% to the left and 2.5% to the right

We can find the \(k\) value if the sample size is 5

qt(0.025, df=5-1)

[1] -2.78

qt(0.975, df=5-1)

[1] 2.78

• Normal distributions are common in nature
• They happen when many things add together
• If the process is Normal, then the sample average is close to the population average
• The confidence interval uses the t-Student distribution