Week 4 Homework

20. Number of surviving offspring

A tree produces many seeds but only 1.1 on average becomes an established mature tree. Probability of survival $p$ is small and $n$ is large, so we can consider a Poisson distribution with $λ = 1.1$ .

We are looking for the $k$ value where the probability of having more than $k$ seeds survive is less than 1%. In other words, we are looking for the value of $k_{x}$ that fits this formula:

0.01 > 1 - \sum_{k = 0}^{k_{x}} P (k)

0.01 > 1 - \sum_{k = 0}^{k_{x}} \frac{λ^{k}}{k!} e^{- λ}

Plot the Poisson distribution $P (k)$ :

```desmos-graph top=0.4 bottom=-0.01 left=-0.1 right=6 --- \poissondist(1.1) ``` We can see that $k=5$ is when we have the first individual probability below 0.01 (1%). So this is probably a good value to check with:

$1 - (P (0) + P (1) + P (2) + P (3) + P (4) + P (5))$

Using estimated values from the plot
$1 - (0.33 + P (1) + P (2) + P (3) + P (4) + P (5))$

24. Generalizing the Skellam Model to perennial plants

We need to add the possibility that the plant survives another year in the same site - no seeds can grow in that site if it lives.

Previously, our assumption was the only sites that are not occupied are the ones that zero seeds reached. Now, there is another independent possibility, the possibility that an occupied site had its plant survive to the next year. At time $t + 1$ , $p x_{t}$ plants survived to the next year. So our total is:

x_{t + 1} = (1 - e^{- α x_{t}}) + (p x_{t})

However, if a site is already occupied by a living plant, it doesn't matter if it is reached by germination. So this equation is counting these sites twice (Probability - Fundamentals#Example Mutually exclusive events). To correct this, we need to multiply the two independent probabilities of seed reaching the site or :

x_{t + 1} = (1 - e^{- α x_{t}}) (1 - p x_{t}) + p x_{t}

(b) At equilibrium, $x_{t + 1} = x_{t} = \hat{x}$ :

\hat{x} = (1 - e^{- α \hat{x}}) (1 - p \hat{x}) + p \hat{x}

If we solve for alpha then plot how this varies with different $p$ values:

α = - \frac{1}{x} \ln (\frac{1 - x}{1 - p x})

In the plot:

The x axis is alpha (number of seeds produced) and y axis is $\hat{x}$ (fraction of sites occupied at equilibrium)
$p = 0$ is purple, then the curves to their left are increasing $p = (0.2, 0.5, 0.8)$

```desmos-graph top=1 bottom=0 left=0 right=5 --- a=0.8 b=0.5 c=0.2 d=0 y = (1-e^{-xy})(1-ay)+ay y = (1-e^{-xy})(1-by)+by y = (1-e^{-xy})(1-cy)+cy y = (1-e^{-xy})(1-dy)+dy ``` Takeaways from this plot: - When $p=0$, there is no chance of the plant surviving, so when the equilibrium number of sites is close to 0 the number of seeds produced is close to 1 - just enough to replace itself once without competitive effect - When $p$ increases, this makes it so the number of seeds produced with no competition lower, since its more likely the plant will survive in the same site and this seed will be wasted. # 25. Coexistence of two species in the Skellam Model

The superior species is the same since it always out competes the inferior one. There is one condition for success - at least one seed must reach the site. So $x_{t + 1} = 1 - P (0 s e e d s)$

x_{t + 1} = 1 - e^{- α x_{t}}

The inferior species has two conditions for success:

No seeds from the superior species can reach the site ( $P (k_{x} = 0) = e^{- β y}$ )
At least one seed from $y$ must reach the site ( $1 - P (k_{y} = 0)$ )

These two are independent so we multiply them:

y_{t + 1} = (1 - e^{- β y_{t}}) e^{- α x_{t}}

At equilibrium $y = y_{t + 1} = \hat{y}$ . So:

\hat{y} = (1 - e^{- β \hat{y}}) e^{- α \hat{x}}

Solving for $α$ in equation one gives

α = - \frac{\ln (1 - \hat{x})}{\hat{x}}

Solving for $β$ in equation two is $β = - \frac{1}{y} \ln (1 - \frac{y}{e x p (- α x)})$

If we sub in the $α$ from the first part, it simplifies to:

β = - \frac{1}{y} \ln (1 - \frac{y}{1 - x})

Now, we can try plotting alpha as a function of x and beta as a function of y

If we plot the corresponding alpha and beta values according to certain $x$ values if we consider that $y \to 0$ ?

(not exactly sure what happens here)

When we consider y as very small, using some property of a logarithm that I don't understand, we can simplify the beta equation to
$β \approx \frac{1}{1 - x}$

26. Host parasitoid system

Imagine the Skellam model, except instead of sites we are talking about hosts for the parasites. Each host can support $B$ parasite larvae. The host species will die after an infection but non-parasitized hosts produce $F$ offpsring each.

Also, let's say that $α$ how many attacks one parasitoid makes per host:

α = \frac{n}{P_{t} H_{t}}

Total number of attacks is given by $n = α P_{t} H_{t}$
One parasitoid makes $α H_{t}$ attacks
Chance of a host being infected is $p = 1 / H_{t}$ - this is very small

Since number of attacks ( $n$ ) is very large and the chance of an individual host being infected ( $p$ ) is small, we can use a Poisson distribution with lambda:

λ = n p = (α P_{t} H_{t}) (\frac{1}{H_{t}}) = α P_{t}

(a) Probability that the host avoids attack

This is the probability that zero parasitoid offspring reach this host. Given the parameters, and assuming $n$ infected hosts and $\frac{n}{H_{t}} = α$ , they might follow a Poisson dist with $λ = P_{t} α$ . The probability of zero infections is wtih $k = 0$ :

P (k = 0) = e^{- P_{t} α}

(b) Two conditions for a host being alive are:

Not getting infected (current number * probability of no infection)
Reproducing ( $F$ )
If the host survives, it can reproduce and make $F$ offspring. So multiplying conditions 1 and 2 gives our output:

H_{t + 1} = H_{t} e^{- α P_{t}} F

For the parasite, it produces $B$ offspring in each infected host $H_{t}$ . It can infect $1 - e^{α P_{t}}$ hosts. So the total is:

P_{t + 1} = H_{t} (1 - e^{- α P_{t}}) B

Bonus: Coin flipping

We flip two fair coins. If the number of heads is $x$ (may be 0, 1, 2), then you get $x^{2}$ euros. Calculate (a) the expected number of heads per round; and (b) the expected gain per round.

Probability - Mean#Example gambling

(a) Number of heads per round:
$E (ξ) = \sum_{k = 0}^{2} x_{k} \cdot P_{k}$

Define $x$ and $k$ for each possibility:
$x_{0} = 0, P_{0} = 0.5 \cdot 0.5 = 0.25$
There is a $0.25$ possibility of getting 0 heads

$x_{1} = 1, P_{1} = (0.5 \cdot 0.5) + (0.5 \cdot 0.5) = 0.5$ :
We add the possibility of getting heads on coin one and not coin 2 plus the inverse. There is a 0.5 probability of 2 heads.

$x_{2} = 2, P_{0} = 0.5 \cdot 0.5 = 0.25$

$E (ξ) = x_{0} P (0 heads) + x_{1} P (1 heads) + x_{2} P (2 heads)$
$E (ξ) = 0 (0.25) + 1 (0.5) + 2 (0.25)$
$E (ξ) = 1$

(b) expected money per round.

Since number of money is $x^{2}$ and we have the same probabilities:
$E (ξ) = x_{0}^{2} P (0 heads) + x_{1}^{2} P (1 heads) + x_{2}^{2} P (2 heads)$
$E (ξ) = 0 (0.25) + 1^{2} (0.5) + 2^{2} (0.25)$
$E (ξ) = 1.5$