Sums and Limits of Coin Throws

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 X 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 p\in [0,1]. We write X\sim Bern(p)

Ex 1 [Bernoulli to Binomial] Let X_i, i=1,...n be IID Bernoulli RVs, let

Show that

Ans 1 Probability of k ones in a row and n-k zeros in a row is p^k (1-p)^{n-k}. The number of sequence with k ones and n-k zeros is

Def [Binomial Distribution] A RV Y has a Binomial distribution, and we write Y\sim Bin(n,p) when

Ex 2 [Bernoulli to Geometric] Consider a sequence of Bernoulli RVs X_1,X_2,.... Let G be the index of the first 1 in this sequence. Show that

Ans 2

Def [Geometric Distribution] A RV G has Geometric Distribution, and we write G \sim Geom(p) if

We now consider some limits of Binomial Random Variables.

Ex 3 [Binomial to Poisson]Consider a sequence of Binomial RVS: Y^n\sim Bin (n, \frac{\lambda}{n}) for some \lambda >0. Show that

Ans 3

Above the term in square brackets goes to one and (1-\lambda/n)^n goes to e^{-\lambda}.

Def [Poisson Distribution] For parameter \lambda>0, a RV N Poisson distribution and we write N\sim Po(\lambda) if

Ex 4 [Geometric to Exponential] Consider a sequence of Geometric RVS: X^n\sim Geom (\frac{\lambda}{n}) for some \lambda >0. 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 Y \sim Bin (2n , \frac{1}{2}) then

Def [Normal Distribution] For mean \mu and variance \sigma^2 we say that Z has a normal distribution and write Z\sim {\mathcal N}(\mu,\sigma^2) when

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

Ex 5 [Binominal to Normal] If Y \sim Bin (2n , \frac{1}{2}) Show that

Ans 6

Cancelling and dividing by n^k gives the required result.

Ex 7 [Continued] Show that

Ans 7 Applying the approximation \left( 1-x \right)= e^{-x +O(x^2)} we have that

Applying this also to the denominator gives the result.

Ex 8 Show, using Stirling’s Approximiation, that, as n\rightarrow \infty,

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 9ac8668cf8418f72ec6d7d9970b9b1c5.pngIn our case the probability corresponds to the sum

After applying substitution x\sqrt{\frac{n}{2}}=k.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s