We explain why certain distributions arise naturally as the limit of coin throws.

- Bernoulli, Binomial Distributions, Geometric Distributions.
- Binomial to Poisson Distribution; Geometric to Exponential; Binomial to Normal.

Bernoulli random variables are just choice tosses: heads or tails; zero or one random variables. You can get surprisingly far – perhaps almost everywhere – in probability by summing and limiting sequences of these random variables.

**Def** [Bernoulli Random Variable] A random variable that takes at most two values is called a Bernoulli random variable. We assume (unless stated otherwise) that these values are zero or one. So

for some . We write

**Ex 1** [Bernoulli to Binomial] Let , be IID Bernoulli RVs, let

Show that

**Ans 1** Probability of ones in a row and zeros in a row is . The number of sequence with ones and zeros is

**Def** [Binomial Distribution] A RV has a Binomial distribution, and we write when

**Ex 2** [Bernoulli to Geometric] Consider a sequence of Bernoulli RVs . Let be the index of the first in this sequence. Show that

**Ans 2**

**Def** [Geometric Distribution] A RV has Geometric Distribution, and we write if

We now consider some limits of Binomial Random Variables.

**Ex 3** [Binomial to Poisson]Consider a sequence of Binomial RVS: for some . Show that

**Ans 3**

Above the term in square brackets goes to one and goes to .

**Def** [Poisson Distribution] For parameter , a RV Poisson distribution and we write if

**Ex 4** [Geometric to Exponential] Consider a sequence of Geometric RVS: for some . Show that

**Ans 4**

We now work to show that the sum of binomial distributions converges to a specific distribution called the normal distribution.

**Thrm** [Binomial to Normal] If then

**Def** [Normal Distribution] For mean and variance we say that has a normal distribution and write when

This is a special case of the central limit theorem, and involves several steps.

**Ex 5** [Binominal to Normal] If Show that

**Ans 6**

Cancelling and dividing by gives the required result.

**Ex 7** [Continued] Show that

**Ans 7** Applying the approximation we have that

Applying this also to the denominator gives the result.

**Ex 8** Show, using Stirling’s Approximiation, that, as ,

**Ans 8** By Stirlings, $latex n! \sim \sqrt{2\pi n}\cdot e^{-n} n^n$,

**Ex 9** [Continued] Argue that Thrm [SL:Bin2Norm] holds i.e. that

**Ans 9** In our case the probability corresponds to the sum

After applying substitution .