1. Suppose people arrive at a bank with poisson rate $$\lambda = 4$$ per hour.
1. What is the probability that 5 people arrive in the first half hour?
2. What is the probability that at least 3 people arrive in the first hour?
Solution
1. $$P(X = 5|\lambda t = 4*0.5)= \dfrac{e^{-\lambda t}(\lambda t)^5}{5!}$$
#pdf of poisson
dpois(5,lambda=4*0.5)
## [1] 0.03608941
1. $$P(X \ge 3|\lambda t = 4) = 1 - P(X \le 3|\lambda t = 4) = 1 - \sum_{i=0}^3 \dfrac{e^{-\lambda t}(\lambda t)^i}{i!}$$\$
#pdf of poisson
1 - ppois(3,lambda=4)
## [1] 0.5665299
1. Patients arrive at the doctor’s office according to Poisson distribution with $$\lambda = 4$$/hour.

1. What is the probability of getting less than or equal to 8 patients within 2 hours?
2. Suppose each arriving patient has 25% chance to bring a person to accompany. There are 20 seats in the waiting room. At least many hours should pass that there is at least 50% probability that the waiting room is filled with patients and their relatives?
1. $$P(X\le 8|\lambda t = 4*2)= \sum_{i=0}^8 \dfrac{e^{-\lambda t}(\lambda t)^i}{i!}$$
#cdf of poisson
ppois(8,lambda=4*2)
## [1] 0.5925473
1. First let's define the problem. Define $$n_p$$ as the number of patients and $$n_c$$ is the number of company. We want $$n_p + n_c \ge 20$$ with probability 50% or higher for a given $$t^*$$. Or to paraphrase, we want $$n_p + n_c \le 19$$ w.p. 50% or lower.

What is $$n_c$$ affected by? $$n_p$$. It is actually a binomial distribution problem. $$P(n_c = i|n_p) = \binom{n_p}{i} (0.5)^i*(0.5)^{n_p-i}$$. It is even better if we use cdf $$P(n_c \le k|n_p) = \sum_{i=0}^{k} \binom{n_p}{i} (0.5)^i*(0.5)^{n_p-i}$$.

We know the arrival of the patients is distributed with poisson. So, $$P(n_p = j|\lambda t^*) = \dfrac{e^{-\lambda t}(\lambda t)^j}{j!}$$. So $$P(j + k \le N) = \sum_{a=0}^j P(n_p = a|\lambda t^*)*P(n_c \le N-a | n_p = a)$$. Remember it is always $$n_c \le n_p$$.

#Let's define a function
calculate_probability<-function(N=19,t_star=1,lambda=4,company_prob=0.25){
#N is the max desired number of patients
the_prob<-0
for(n_p in 0:N){
the_prob <- the_prob + dpois(n_p,lambda=lambda*t_star)*pbinom(q=min(N-n_p,n_p),size=n_p,prob=company_prob)
}

return(the_prob)

}

#Try different t_stars so probability is below 0.5
calculate_probability(t_star=3)
## [1] 0.8380567
calculate_probability(t_star=3.3)
## [1] 0.7443518
calculate_probability(t_star=3.955)
## [1] 0.49914
1. Suppose the the pdf of a random variable $$x$$ is $$f(x) = \dfrac{a}{(1-x)^{1/3}}$$ for $$0 < x < 1$$ and $$0$$ for other values of x. (Note: There was a typo in the in-class exercise saying the interval is $$0 < x < 1$$)

1. Find the constant $$a$$.
2. Find cdf of F(X < 3/4).

Integral of $$f(x)$$ should be 1. So $$\int_0^1 f(x) dx = a*(-3/2(1-x)^{2/3})|_0^1 = 3a/2$$. Then $$a = 2/3$$.

Use the same integral $$\int_0^{3/4} f(x) dx = -(1-x)^2/3|_0^{3/4} = 1 - (1/4)^{2/3} = 0.603$$

1. Let $$X$$ and $$Y$$ be the random variables and $$f(x,y)$$ is the probability density function of the joint distribution. Suppose $$f(x,y) = a(\dfrac{5x}{7} + \dfrac{9y^3}{2})$$ if $$0<x<2$$ and $$-1<y<1$$ (0 otherwise).

1. Find $$a$$.
2. Find the marginal distribution of $$y$$ ($$h(y)$$) and $$h(y<0.5)$$.
3. Find the conditional distribution of $$f(y|x)$$.

a) $$\int^1_{-1} \int^2_0 a(\dfrac{5x}{7} + \dfrac{9y^3}{2}) dx dy = \int^1_{-1} a(\dfrac{5x^2}{14} + \dfrac{9xy^3}{2})dy|^2_0 = \int^1_{-1} a(\dfrac{10}{7} + 9y^3) dy$$. (This is also $$h(y)$$ if $$a$$ is known.) $$a(\dfrac{10y}{7} + \dfrac{9y^4}{4})|^1_{-1} = a(20/7)$$. In order to be a distribution it should be equal to 1. So $$a = 7/20$$.

b) As given in (a) $$h(y) = 7/20*(\dfrac{10}{7} + 9y^3)$$. So $$h(y<0.5) = \int^{0.5}_{-1} 7/20*(\dfrac{10}{7} + 9y^3) dy = 7/20*(\dfrac{10y}{7} + \dfrac{9y^4}{4})|^{0.5}_{-1} = 0.0335$$

c) We need to find $$g(x) = \int^1_{-1} 7/20(\dfrac{5x}{7} + \dfrac{9y^3}{2})dy = x/2$$. $$f(y|x) = f(x,y)/g(x)$$.