# Plot the Poisson CDF with the Standard Normal Distribution CDF

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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')

##### 0 Comments

### Answers (3)

VBBV
on 7 Jan 2022

Edited: VBBV
on 7 Jan 2022

x3 = -3:0.1:3

Use the same intervals as x1

##### 1 Comment

John D'Errico
on 8 Jan 2022

John D'Errico
on 8 Jan 2022

Edited: John D'Errico
on 8 Jan 2022

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...

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.

##### 0 Comments

Torsten
on 8 Jan 2022

Edited: Torsten
on 8 Jan 2022

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

##### 0 Comments

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