Week 5 Homework

11. Luria-Delbrück fluctuation test

Probability - Fundamentals
For estimating mutation rates in bacteria.

Each sample has divided $10^{8}$ times / mL. Since we are using a 0.1mL sample, we are considering that there are $10^{7}$ mutations in our sample . If any resistant bacteria is in the sample it will grow on the toxic plate.

In 20 trials, 11 had a mutation frequency of $0 / 10^{7}$ , while the remainder had at least one.

Probability of mutation in one division is $μ$ .
Probability of no mutations in one division is $1 - μ$
Probability of no mutations in $10^{7}$ divisions is $(1 - μ)^{10^{7}}$
This must match the observed frequency of no mutations in $n = 10^{8}$ divisions = $11 / 20$

P (0) = (1 - μ)^{n}

P (0)^{n^{- 1}} = 1 - μ

μ = 1 - P (0)^{n^{- 1}}

μ = 1 - {(\frac{11}{20})}^{10^{- 7}} = 5.98 \times 10^{- 8}

Number of divisions

I don't fully understand why we are considering the $10^{7}$ divisions that are in the sample instead of the total number of divisions in the test tube. I would think that a mutation that happened anywhere in the population could end up in our sample if it is representative, so we should use the total number of bacteria in the sample? This is incorrect but I don't completely understand why.

Poisson Distributions

We can also do this with the Poisson distribution with $λ = n μ$ , since we have many bacteria, a very low probability and two outcomes (mutation or no mutation). So we can use Poisson distribution with $k = 0$ :

P (0) = \frac{(n μ)^{0}}{0!} e^{- n μ}

P (0) = e^{- n μ}

\ln (P (0)) = - n μ

μ = - \frac{\ln (P (0))}{n} = - 10^{- 7} \ln (\frac{11}{20}) = 5.98 \times 10^{- 8}

29. Nestling feeding variance

Probability - Variance
The mother brings to the nest on average:
${\bar{ξ}}_{1} = 24$
${σ_{1}}^{2} = 2.3$
The father brings:
${\bar{ξ}}_{2} = 21$
${σ_{2}}^{2} = 3.1$

Assuming these are independent, the total mean amount of food brought is:
$E ({\bar{ξ}}_{1} + {\bar{ξ}}_{2}) = {\bar{ξ}}_{1} + {\bar{ξ}}_{2} = 45 g$
The variance works the same way:
$V (ξ_{1} + ξ_{2}) = σ_{1}^{2} + σ_{2}^{2} = 5.4 g$

(b) If there are 9 chicks in the nest and each gets the same amount of food, and these are independent, we can use the coefficient rule to simply divide by 9 for the mean:
$45 / 9 = 5 g$
But for the variance, we need to square the coefficient:
$(1 / 9)^{2} \cdot 5.4 = 0.667 g$

(c) Perhaps only one parent can be foraging while the other guards the nest. In this case there is a negative correlation between the amount each can forage. In this case, we expect that the total variance would be lower than the given sum. I'm not sure about the coefficient, this wasn't in the notes.

31. Expectation from dice rolls

Probability - Mean
$η$ is the average of 10 random numbers from casting a (assuming 6-sided) die.

To make it clear that this is the average let's call it $\bar{η}$ .

E (\bar{η}) = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = \frac{7}{2} = 3.5

(Expectation doesn't depend on sample size)
Probability - Variance

True variance $V (η)$

Since $E (η) = E (\bar{η})$ , we can calculate the true variance of the theoretical infinite sample based on the expectation of the sample average:

$V (η) = E ({\bar{η}}^{2}) - [E (\bar{η})]^{2}$
$[E (\bar{η})]^{2} = {3.5}^{2} = 12.25$

E ({\bar{η}}^{2}) = \frac{1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2}}{6} = \frac{91}{6} = 15.16 . .

$V (η) = 15.16 - 12.25 = 2.91$
Then, the variance of the sample average, which depends on the sample size, is given by:

V (\bar{η}) = \frac{1}{n} V (η) = \frac{1}{10} \cdot 2.91 = 0.291

32 Variance of independent variables

Probability - Variance#Sum of variances
(i) The variance of the difference $ξ - η$ if they are independent is just $V (ξ) + V (- η)$

$η$ has a coefficient of -1, so we can take thisoutside the term and square it, giving us $V (ξ) + V (η)$

Probability - Variance#Coefficient with variance
(ii) If we use this constant $a$ , we apply the same rule, but since its a constant it has a variance of zero:
$V (ξ + a) = V (ξ)$
Same for minus.

If you think of it on a graph this makes sense. For example for height: if we subtract 100cm from everyone's height the variance is the same.

34. Variance vs the "accuracy" of measurement

Suppose we would like to measure the frequency $p$ of a certain trait (genotype, disorder, etc.) in a population. To this end, we take a sample of $n$ individuals and count the number $ξ$ of those in the sample who have the trait.

(a) Describe the conditions under which $ξ$ is a binomially distributed random variable

The probability of having $ξ$ individuals with a certain trait in a population of $n$ with frequency $p$ is given by the binomial distribution:

P (ξ) = (\binom{n}{ξ}) p^{ξ} (1 - p)^{n - ξ}

(b) One has the intuitive feeling that with larger sample size $n$ , variation should somehow dampen and our measurement should become more accurate. Does the variance of $ξ$ become smaller as $n$ increases? Or the standard deviation, $D (ξ) = \sqrt{V (ξ)}$ ? Or the coefficient of variation, $c = D (ξ) / E (ξ)$ ? Or the variance of the estimated frequency, $V (ξ / n)$ ?

Variance in binomial distribution:
$V (ξ) = E (ξ^{2}) - [E (ξ)]^{2}$
We know $E (ξ) = n p$ , as proved in the lecture (Expectation of Binomial Distribution). Quick recap:

Recap of expectation of binomial dist

Consider that the distribution $ξ$ is made of the sum of all the $n$ individual random variables $η_{i}$ .

E (ξ) = E (η_{1}) + E (η_{2}) + . . .

For the expectation of each individual response, we just multiply each outcome by its probability. Our outcomes are 1 and 0 and their probabilities are $p$ and $1 - p$ :

E (η_{1}) = 1 \cdot p + 0 \cdot (1 - p) = p

Therefore, our expectation of the binomial distribution is the sum of all of the $n$ responses, which are always $p$ . So, $E (ξ) = n p$ for a binomial distribution.

Similarly to above, let's calculate the variance of one response $V (η_{1})$

V (η_{1}) = E (η_{1}^{2}) - [E (η_{1})]^{2}

As shown above, $E (η_{i}) = 1 \cdot p + 0 \cdot (1 - p)$ .
So $E (η_{i}^{2}) = 1^{2} \cdot p + 0^{2} \cdot (1 - p) = p$

V (η_{i}) = p - p^{2} = p (1 - p)

Since the sum of the variance is the variance of the sum, and $ξ$ is the sum of $η_{1}, η_{2}, . . .$
$V (ξ) = V (η_{1}) + . . . + V (η_{n})$

Each term $V (η_{i})$ is $p (1 - p)$ . There are $n$ terms, so the total variance can be given by:

V (ξ) = n p (1 - p)

Therefore, the variance increases with increasing $n$ , unlike the expectation.

Similarly, the standard deviation ( $\sqrt{V (ξ)}$ ) also decreases since it is directly proportional to $V$ .

The coefficient of variation $c$ :
$c = D (ξ) / E (ξ)$

c = \frac{\sqrt{n p (1 - p)}}{n p}

c = \frac{\sqrt{n p} \sqrt{(1 - p)}}{\sqrt{n p} \sqrt{n p}}

c = \sqrt{\frac{(1 - p)}{n p}}

In this case $c$ increases with $n$ since its inversely proportional to the mean (a measure relative to typical values)

I would expect the variance of estimated frequency should decrease with $n$ . If $n$ is independent we can extract it:

V (ξ / n) = \frac{1}{n^{2}} V (ξ)

V (ξ / n) = \frac{1}{n^{2}} (n p (1 - p))

V (ξ / n) = \frac{p (1 - p)}{n}

Therefore, yes this does get smaller with increasing $n$ .