Distributions Continued

Geometric Series, Cont’d

A reminder about Geometric Series

Let T₁ = Number of trials until the first (or next) “success”

Distribution?

P(T₁ = K) = ?, K = 1,2,… , ∞

$$P(T_1=k)=p{q}^{k-1},k=1,2,..,\infty$$

To make this easier, the expected value is just.

$$E(T_1)=\frac{1}{p}$$

When is the expected next time that something will happen given the probability?

For variance, it’s a little bit more non-sensible.

$$Var(T_1)=\frac{8}{p^2}$$

---

Example One

• How shall we model this?

Let’s assume each year is a trial, and each trial will have a a max wind which either exceeds, or doesn’t the “50-year windspeed” (E(T₁) = 50)

Let X_i be 1 if our windspeed exceeds the 50 year value, and 0 if it does not.

$$P=P(x_i=1)=\frac{1}{50}=\frac{1}{E(T_1)}=0.02$$

Note

Often times, you’ll be given an expected time, and not directly a probability. You’ll just have to flip it like I did above.

A) We want P(T₁ = 5) (F, F, F, F, S)

$$P(T_1=5)=p{q}^{5-1}=p{q}^4=(0.02)(0.98{)}^4=\underline{0.018}$$

B) We want P(N₅ ≥ 1 ) = P(T₁ ≤ 5 ) (F,S,F,F,F)

---

Negative Binomial

Let T_k = Number of trials until the k^th “success”

Distribution?

P(T_k = m) = ?, M = k, k+1,… , ∞

$$P[T_k=m]=\left(\genfrac{}{}{0ex}{}{m-1}{k-1}\right)p^k{q}^{m-k},m=k,k+1,...$$

This is pretty similar to Geometric! If K =1, then it really is the same. It’s a generalization.

---

M = Number of Trials K = Number of Successes in M Trials

And the variance of T_k is

$$Var(T_k)=Var(T_{11}+T_{12}+...+T_{1k})=Var(T_{11})+Var(T_{12})..$$ $$Var(T_k)=\frac{q}{p^2}+\frac{q}{p^2}+...=\frac{kq}{p^2}$$

And the expected value is

$$E(T_k)=E(T_{1}1)+E(T_{1}2)+...+E(T_{1}k)$$ $$E(T_k)=\frac{1}{p}+\frac{1}{p}+...+\frac{1}{p}=\frac{k}{p}$$

---

Example Two

In the transmission tower problem from above, what is the probability that the second 50 year wind will occur in the 5th year of service?

We want P(T₂ = 5)

$$P(T_2=5)=\left(\genfrac{}{}{0ex}{}{4}{1}\right)p^2{q}^3$$ $$P(T_2=5)=\left(\genfrac{}{}{0ex}{}{4}{1}\right)(0.02{)}^2(0.98{)}^3=\underline{0.0015}$$

---

Now, let’s do something fun.

The Rate of Success, Poisson Distributions

Lets let every instant in time become a Bernoulli Trial. Now, “successes” can occur at any instant

Problems: Let N_n equal the number of “successes” in time t (?) N > ∞ on any time interval

$$E(N_n)=np\to \infty \text{unless}p\to 0$$

If P > 0

We need to replace P with a mean rate of “successes”, λ (number per unit time)

Let N_t = Number of “successes” in time T (discrete)

Poisson Distribution

$$P(N_t=k)=\frac{(\lambda t{)}^k}{k!}{e}^{-\lambda t},t\ge 0$$

This is the analog to the binomial distribution (N_n = number of successes )

and of course…

Expected Value

$$E(N_t)=\lambda t$$

Variance

$$Var(N_t)=\lambda t\to \sqrt{\lambda t}$$

---

Summary of Statistics Distributions