# Difference between revisions of "binomial Probability Monte Carlo Sampling June 2 2009"

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You are given two independent Binomial distributions with probabilities <math>\displaystyle p_1\text{, }p_2</math>. Using a Monte Carlo simulation, approximate the value of <math>\displaystyle \delta</math>, where <math>\displaystyle \delta = p_1 - p_2</math>.<br> | You are given two independent Binomial distributions with probabilities <math>\displaystyle p_1\text{, }p_2</math>. Using a Monte Carlo simulation, approximate the value of <math>\displaystyle \delta</math>, where <math>\displaystyle \delta = p_1 - p_2</math>.<br> |

## Latest revision as of 08:45, 30 August 2017

##### Example 1:

You are given two independent Binomial distributions with probabilities [math]\displaystyle p_1\text{, }p_2[/math]. Using a Monte Carlo simulation, approximate the value of [math]\displaystyle \delta[/math], where [math]\displaystyle \delta = p_1 - p_2[/math].

- [math]\displaystyle X \sim BIN(n, p_1)[/math]; [math]\displaystyle Y \sim BIN(n, p_2)[/math]; [math]\displaystyle \delta = p_1 - p_2[/math]

So [math]\displaystyle f(p_1, p_2 | x,y) = \frac{f(x, y|p_1, p_2)*f(p_1,p_2)}{f(x,y)}[/math] where [math]\displaystyle f(x,y)[/math] is a flat distribution and the expected value of [math]\displaystyle \delta[/math] is as follows:

- [math]\displaystyle \hat{\delta} = \int\int\delta f(p_1,p_2|X,Y)\,dp_1dp_2[/math]

Since X, Y are independent, we can split the conditional probability distribution:

- [math]\displaystyle f(p_1,p_2|X,Y) \propto f(p_1|X)f(p_2|Y)[/math]

We need to find conditional distribution functions for [math]\displaystyle p_1, p_2[/math] to draw samples from. In order to get a distribution for the probability 'p' of a Binomial, we have to divide the Binomial distribution by n. This new distribution has the same shape as the original, but is scaled. A Beta distribution is a suitable approximation. Let

- [math]\displaystyle f(p_1 | X) \sim \text{Beta}(x+1, n-x+1)[/math] and [math]\displaystyle f(p_2 | Y) \sim \text{Beta}(y+1, n-y+1)[/math], where
- [math]\displaystyle \text{Beta}(\alpha,\beta) = \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}p^{\alpha-1}(1-p)^{\beta-1}[/math]

**Process:**

- Draw samples for [math]\displaystyle p_1[/math] and [math]\displaystyle p_2[/math]: [math]\displaystyle (p_1,p_2)^{(1)}[/math], [math]\displaystyle (p_1,p_2)^{(2)}[/math], ..., [math]\displaystyle (p_1,p_2)^{(n)}[/math];
- Compute [math]\displaystyle \delta = p_1 - p_2[/math] in order to get n values for [math]\displaystyle \delta[/math];
- [math]\displaystyle \hat{\delta}=\frac{\displaystyle\sum_{\forall i}\delta^{(i)}}{N}[/math].

**Matlab Code:**

- The Matlab code for recreating the above example is as follows:

n=100; %number of trials for X m=100; %number of trials for Y x=80; %number of successes for X trials y=60; %number of successes for y trials p1=betarnd(x+1, n-x+1, 1, 1000); p2=betarnd(y+1, m-y+1, 1, 1000); delta=p1-p2; mean(delta);

The mean in this example is given by 0.1938.

A 95% confidence interval for [math]\delta[/math] is represented by the interval between the 2.5% and 97.5% quantiles which covers 95% of the probability distribution. In Matlab, this can be calculated as follows:

q1=quantile(delta,0.025); q2=quantile(delta,0.975);

The interval is approximately [math] 95% CI \approx (0.06606, 0.32204) [/math]

Note: In this case, we can also find [math]E(\delta)[/math] analytically since [math]E(\delta) = E(p_1 - p_2) = E(p_1) - E(p_2) = \frac{x+1}{n+2} - \frac{y+1}{m+2} \approx 0.1961 [/math]. Compare this with the maximum likelihood estimate for [math]\delta[/math]: [math]\frac{x}{n} - \frac{y}{m} = 0.2[/math].

##### Example 2:

We conduct an experiment by giving rats one of ten possible doses of a drug, where each subsequent dose is more lethal than the previous one:

- [math]\displaystyle x_1\lt x_2\lt ...\lt x_{10}[/math]

For each dose [math]\displaystyle x_i[/math] we test n rats and observe [math]\displaystyle Y_i[/math], the number of rats that survive. Therefore,

- [math]\displaystyle Y_i \sim~ BIN(n, p_i)[/math]

.

We can assume that the probability of death grows with the concentration of drug given, i.e. [math]\displaystyle p_1\lt p_2\lt ...\lt p_{10}[/math]. Estimate the dose at which the animals have at least 50% chance of dying.

- Let [math]\displaystyle \delta=x_j[/math] where [math]\displaystyle j=min\{i|p_i\geq0.5\}[/math]
- We are interested in [math]\displaystyle \delta[/math] since any higher concentrations are known to have a higher death rate.

**Solving this analytically is difficult:**

- [math]\displaystyle \delta = g(p_1, p_2, ..., p_{10})[/math] where g is an unknown function
- [math]\displaystyle \hat{\delta} = \int \int..\int_A \delta f(p_1,p_2,...,p_{10}|Y_1,Y_2,...,Y_{10})\,dp_1dp_2...dp_{10}[/math]

- where [math]\displaystyle A=\{(p_1,p_2,...,p_{10})|p_1\leq p_2\leq ...\leq p_{10} \}[/math]

- where [math]\displaystyle A=\{(p_1,p_2,...,p_{10})|p_1\leq p_2\leq ...\leq p_{10} \}[/math]

**Process: Monte Carlo**

We assume that

- Draw [math]\displaystyle p_i \sim~ BETA(y_i+1, n-y_i+1)[/math]
- Keep sample only if it satisfies [math]\displaystyle p_1\leq p_2\leq ...\leq p_{10}[/math], otherwise discard and try again.
- Compute [math]\displaystyle \delta[/math] by finding the first [math]\displaystyle p_i[/math] sample with over 50% deaths.
- Repeat process n times to get n estimates for [math]\displaystyle \delta_1, \delta_2, ..., \delta_N [/math].
- [math]\displaystyle \bar{\delta} = \frac{\displaystyle\sum_{\forall i} \delta_i}{N}[/math].