Question

The value of the R function "mean" applied to a vector $\mathbf{v}=(v_1,v_2,...v_n)$ is the arithmentic mean of the vector: $\bar{v}=\frac{1}{n} \sum_{i=1}^nv_i$. The value of the R function "var" applied to the vector $\mathbf{v}$ equals $\frac{1}{n-1} \sum_{i=1}^n(v_i-\bar{v})^2$, a measure of how much the values differ from the mean. For $\lambda\in\{4,25,100\}$, create samples of size 100,000 from the Poisson distribution with parameter $\Lambda$ and the Normal distribution with mean equal to $\lambda$ and sd equal to $\sqrt{\Lambda}$. Please compare the values of "mean", "var", "max", and "min" for the Poisson and the Normal samples corresponding to each $\lambda$.(5 points) ### 3.b. For the values of $\lambda$ and for integers $x$ with the lower bound of O and the upper bounds of 15, 60, and 150, respectively, provide visualizations comparing the probability of an outcome equal to $x$ under $Poisson(\Lambda)$ and a value in $(x-.5,x+.5) $ under the Normal distribution with $\mu=\lambda$ and $\sigma^2=\lambda$. Also, for each $\lambda$, give the sum of the absolute differences of the two probabilities for all the values of $x$ assessed. (5 points) "*'{r}

Answer 1

R code with comments

#set the random seed set. seed(123) #set the sample size n<- 100000 #loop thorugh for different values of lambda for (lambda in c(4,25,100)) { #draw n sample from poisson(lambda) v.p<-rpois (n, lambda) #draw n samples from normal (lambda, lambda) v.n<-rnorm(n, lambda,sqrt(lambda)) print (sprintf('for lambda=%g-->', lambda)) cat (sprintf('\t the mean for poisson samples=%.4f, for Normal samples: %.4f\n',mean(v.p), mean(v.n)) cat (sprintf('\t the variance for Poisson samples=%.4f, for normal samples: %.4f\n', var (v.p),var (v.n))) cat (sprintf('\t the max for Poisson samples=%. 4f, for normal samples: %.4f\n',max (v.p),max(v.n))) cat (sprintf('\t the min for Poisson samples=%. 4f, for Normal samples: %.4f\n',min(v.p),min(v.n)))

get this

[1] "for lambda=4 -->" the mean for Poisson samples=3.9973, for normal samples: 4.0071 the variance for Poisson samples=3. 9815, for Normal samples: 4.0161 the max for Poisson samples=15.0000, for normal samples: 12.4122 the min for Poisson samples=0.0000, for normal samples: -4.7642 [1] "for lambda=25-->" the mean for Poisson samples=24.9949, for Normal samples: 24.9921 the variance for Poisson samples=24.9313, for normal samples: 25.0205 the max for Poisson samples=48.0000, for normal samples: 47.2735 the min for Poisson samples=7.0000, for normal samples: 3.0824 [1] "for lambda=100-->" the mean for Poisson samples=99.9399, for normal samples: 99.9833 the variance for Poisson samples=100.0834, for normal samples: 99.5701 the max for Poisson samples=147.0000, for normal samples: 142.4003 the min for Poisson samples=52.0000, for normal samples: 56.6652

We can see that the Poisson and normal samples have similar sample means and sample variances for all the values of lambda.

However the minimum and maximum sample values are not similar.

3.b) R code with comments

#set the sample size n<- 100000 #make way for 3 plots par (mfrow=C(3,1)) #loop thorugh for different values of lambda for (lambda in c(4,25,100){ for (ub in c(15,60,150)) { #set the values of x in the interval o to ub X<- 0:ub #calculate the poisson probabilities for x=0:ub prob. p<-dpois (x, lambda) #calculate the norm probabilities for x-0.5,x+0.5 prob.n<-pnorm(X+0.5, lambda,sqrt(lambda))-pnorm (X-0.5, lambda,sqrt(1ambda)) #plot the poisson probabilities plot(x, prob.p, type="1",ylab="probability") lines (x, prob.n, col="red") legend ("topright",c("Poisson", "Normal"), 1ty=1,col=C(1, "red")) #calculate the sum of absolute differences sad<- sum(abs (prob. p-prob.n)) print (sprintf('for lambda=%g-->', lambda)) cat (sprintf('\t the sum of absolute differences is %.4f\n', sad))

get this graph

probability 0.00 0.03 probability 0.00 0.06 probability 0.00 0.15 FO 50 100 100 100 150 Normal Poisson 150 Normal Poisson 150 Normal Poisson

and get these

[1] "for lambda=4 -->" the sum of absolute differences is 0.1178 [1] "for lambda=25-->" the sum of absolute differences is 0.0505 [1] "for lambda=100-->" the sum of absolute differences is 0.0252

The normal approximation to Poisson distribution improves with the increase in value of lambda