Question

Recall that a discrete random variable X has Poisson distribution with parameter λ if the probability mass function of X

Recall that a discrete random variable X has Poisson distribution with parameter λ if the probability mass function of X is r E 0,1,2,...) This distribution is often used to model the number of events which will occur in a given time span, given that λ such events occur on average a) Prove by direct computation that the mean of a Poisson randon variable with parameter λ is simply b) Let X1 ~ Poisson(A1) and X2 ~ Poisson(A2) be independent. Prove that the ra ndom variable X1+X2 is also Poisson with parameter λ1 + λ2 Hint 1: The Persian mathematician Al-Karaji was the first to write down what is today known as the binomial formula: if n e [0,1,2,...), thern rt You will likely find the binomial formula useful in your proof Hint 2: For X, + X2 to be r. we can have X1-0 and X2-2, or X,-1 and X2 = æ-1. or X,-2 and X2- x- 2, and so on. How do we compute P(X1 + X2-x)? c) Suppose that the average number of accidents on the I-5 per day is 5, and that the average number on the I-8 is 11. Assume that the number of accidents on each are independent Poisson random variables What is the probability that the total number of accidents in a day between both expressways is less than or equal to 9? You may use a computer to calculate a decimal answer. For example, scipy.stats.poisson might be useful. But if you use a computer, give the code that you used

Answer 1

$(a)~E(X)=\sum_{x=0}^\infty xe^{-\lambda}\frac{\lambda^x}{x!}=\lambda e^{-\lambda}\sum_{x=1}^\infty\frac{\lambda^{x-1}}{(x-1)!}=\lambda e^{-\lambda}e^\lambda=\lambda\\ (b)~P(Y=X_1+X_2=y)=P\left(\cup_{x=0}^y\{X_1=x,X_2=y-x\} \right )\\ =\sum_{x=0}^yP(X_1=x,X_2=y-x)~Since~\{X_1=x,X_2=y-x\},~x=0(1)y~are~disjoint~events\\ =\sum_{x=0}^yP(X_1=x)P(X_2=y-x),~since~X_1~and~X_2~are~independent.\\ =\sum_{x=0}^ye^{-\lambda_1}\frac{\lambda_1^x}{x!}\times e^{-\lambda_2}\frac{\lambda_2^{y-x}}{(y-x)!}=\frac{e^{-\lambda_1-\lambda_2}}{y!}\sum_{x=0}^y\binom{y}{x}\lambda_1^x\lambda_2^{y-x_1}\\ ==\frac{e^{-\lambda_1-\lambda_2}}{y!}(\lambda_1+\lambda_2)^y,~y=0,1,2,...\\ =0~otherwise\\ Hence~Y\sim Poi(\lambda_1+\lambda_2).\\ (c)~Let~X_1=no.~of~accident~on~the~I-5~day\sim Poi(5)\\ X_2=no.~of~accident~on~the~I-8~day\sim Poi(11)\\ Since~X_1~and~X_2~are~independent~then~Y=X_1+X_2\sim Poi(16).\\ P(Y\leq 9)=\sum_{y=0}^9\frac{e^{-16}(16)^y}{y!}$

R code:

y=0:9
p=exp(-16)*(16)^y/factorial(y)
round(sum(p),4)

Output:

round(sum(p),4)
[1] 0.0433
Answer: 0.0433.