Plot the Poisson CDF with the Standard Normal Distribution CDF

Hello all, I have a question on plotting the Poisson cdf together with the standard normal distribution cdf. Below is the task description:

Let X be a random variable with and be the normalized version of X.

The following figure shows the cdf of for some and the cdf Φ of the standard normal distribution.

I have tried my best to plot as same as the given figure but still I am not able to get it looks similiar as the given figure. Below is my result:

May I know what do I miss? Below is my code:

pe = makedist('Normal')

x1 = -3:.1:3;

p = cdf(pe,x1);

%plot(x1,p)

x3 = -3:3;

y3 = poisscdf(x3,1);

figure

stairs(x3,y3)

hold on

plot(x1,p)

xlabel('Observation')

ylabel('Cumulative Probability')

Answer 1

[John D'Errico]

Note that a Poisson random variable will ALWAYS be a non-negative number. It CANNOT have mass on the negative end of the real line. So the plot that you show, with a supposedly Poisson random variable that goes negative? It cannot exist. (At least not for a standard Poisson.)

Perhaps the intent is to have a Poisson random variable that has been transformed into the range of a normal distribution.

For example, Poisson distributions with large Poisson parameters will tend to look very normally distributed. (Not difficult to prove as I recall.)

fplot(@(x) poisscdf(x,50),[0,100])

So a Poisson CDF that does look quite normal. A quick glance at Wikipedia...

https://en.wikipedia.org/wiki/Poisson_distribution

tells me that for a Poisson distribution with parameter lamnda, the mean will be lambda, as well as the variance. So we can simply do this:

lambda = 30;
fplot(@(x) poisscdf(x,lambda),[0,2*lambda])
hold on
fplot(@(x) normcdf(x,lambda,sqrt(lambda)))
grid on
hold off

s you should see, the two curves nearly overlay on top of each other.

In your figure, the Poisson was apparently transformed, via a transformation to look like a normal.

lambda = 30;
fplot(@(x) poisscdf(lambda + x*sqrt(lambda),lambda),[-3,3])
hold on
fplot(@(x) normcdf(x),[-3,3])
grid on
hold off

However, that is NOT a Poisson distibution because it is shown to have mass for negative x. It is derived from one.

Answer 2

[Torsten]

I suspect the graphics shows that the arithmetic mean of normalized independent Poisson random variables converges in distribution to the standard normal distribution.

You can see it from the following plot:

function main
  lambda = 3;
  m = 2000;   % number of RVs
  n = 3000;   % number of samples 
  a = poissrnd(lambda,m,n);
  for i=1:n
    b(i) = (sum(a(:,i))/m-lambda)/sqrt(lambda/m);  % normalized arithmetic mean of Poisson RVs
  end
  cdfplot(b); % plot empirical cdf of b
  hold on
  x = linspace(min(b),max(b));
  plot(x,normcfd(x,0,1))   % compare with standard normal distribution
end