How to generate random variables in C that conform to a Poisson distribution?

1) In the Poisson distribution, when finding the value of x, when p(x=k) is the maximum, set it to pxmax

At this time, this is the same as solving for f(x)=lamda k k!There is a maximum value when k is taken, although this problem is somewhat difficult, but in the program it is possible to go through a loop to get the integer independent variable xmax when f(x) is taken as the maximum.

2) Through iteration, the random number on the 0-1 interval is continuously generated, when the random number = c ) x ++while ( b >= c );

return x;

by Poisson distribution formula.

You can roll out the relationship between the anterior and post.

In this way, the K+1 item can be rolled out from item k. Now, let's analyze how to calculate random numbers that conform to a Poisson distribution. For the integer k from 1 to positive infinity, rand() a decimal place each time, if less than p(x=k), k is output, and k is a value that conforms to the Poisson distribution.

Repeat the above steps n times to obtain n random numbers that conform to the Poisson distribution. pseudo-** form: p = exp(-lamda); for k = 0 ->infinity randvalue = rand(); if( randvalue < p) cout = 3 * lamda) to prevent the case that cannot be found"---");}" : "+k);

Is it to find the probability of a Poisson distribution?

I wrote about finding the probability of a Poisson distribution.

#include

#include

#define e

void main()

int k,i,re=1;

float u,p;

scanf("%f %d",&u,&k);

for(i=1;i<=k;i++)

re=re*i;

p=(float)pow(e,-u)*pow(u,k)/re;

printf("p=%f",p);

Step 1: Generate a lot of (0,1) evenly distributed random numbers (you can look up the table, but the general software can be directly adjusted) set to x1, x2, x3, x4, x5...

Step 2: Assuming that the Poisson distribution parameter to be simulated is , calculate e (-Step 3: Take x1* x2*xk >= e^(-x1* x2* .x(k+1)

k is the first random number generated, and then remove the k+1 numbers used above, and repeat the above steps.

For example, x1*x2 >= e (-but x1*x2*x3 < e (-then the first random number is 2, and then multiply by x4 and repeat the above steps to produce the second random number.

#include

#include

#include

#include

double rand()

int prand(double n)

while(t >= n);

return y;

void main()}

I think this statement of yours is very strange in terms of probability.

If you want to generate it:

n = poissrnd(5,5,4)/10n =

See how this program handles: function x=poisondist(x0,lamda,n)format long; x=zeros(n,1);for i=1;n; b=1; tol=1; k=0; while tol==1 r=mixmod(x0,10,1); b=b*r(10); if b

Zero-means unit variance is not correlated with a sequence of measurable random variables.

normrnd(0,1,1,n)

0 is the mean. The first 1 standard deviation.

n is the sequence length, or you can also normrnd(0,1) claim one at a time.

The zero-mean variance is an uncorrelated white noise sequence.

Variance》 sigma=

normrnd(0, ,1, n)

Same as above.