CODEWITHSUNDEEP
rpoisson
clattenburg cake
clattenburg cake

Reputation: 1222

Poisson Table in R

I am trying to generate a Poisson Table in R for two events, one with mean 1.5 (lambda1) and the other with mean 1.25 (lambda2). I would like to generate the probabilities in both cases for x=0 to x=7+ (7 or more). This is probably quite simple but I can't seem to figure out how to do it! I've managed to create a data frame for the table but I don't really know how to input the parameters as I've never written a function before: 

name <- c("0","1","2","3","4","5","6","7+")
zero <- mat.or.vec(8,1)
C <- data.frame(row.names=name,
                "0"=zero,
                "1"=zero,
                "2"=zero,
                "3"=zero,
                "4"=zero,
                "5"=zero,
                "6"=zero,
                "7+"=zero)

I am guessing I will need some "For" loops and will involve dpois(x,lambda1) at some point. Can somebody help please?

Upvotes: 1

Views: 1164

Answers (1)

Peyton
Peyton

Reputation: 7396

I'm assuming these events are independent. Here's one way to generate a table of the joint PMF.

First, here are the names you've defined, along with the lambdas:

name <- c("0","1","2","3","4","5","6","7+")
lambda1 <- 1.5
lambda2 <- 1.25

We can get the marginal probabilities for 0-6 by using dpois, and the marginal probability for 7+ using ppois and lower.tail=FALSE:

p.x <- c(dpois(0:6, lambda1), ppois(7, lambda1, lower.tail=FALSE))
p.y <- c(dpois(0:6, lambda2), ppois(7, lambda2, lower.tail=FALSE))

An even better way might be to create a function that does this given any lambda.

Then you just take the outer product (really, the same thing you would do by hand, outside of R) and set the names:

p.xy <- outer(p.x, p.y)
rownames(p.xy) <- colnames(p.xy) <- name

Now you're done:

              0            1            2            3            4            5
0  6.392786e-02 7.990983e-02 4.994364e-02 2.080985e-02 6.503078e-03 1.625770e-03
1  9.589179e-02 1.198647e-01 7.491546e-02 3.121478e-02 9.754617e-03 2.438654e-03
2  7.191884e-02 8.989855e-02 5.618660e-02 2.341108e-02 7.315963e-03 1.828991e-03
3  3.595942e-02 4.494928e-02 2.809330e-02 1.170554e-02 3.657982e-03 9.144954e-04
4  1.348478e-02 1.685598e-02 1.053499e-02 4.389578e-03 1.371743e-03 3.429358e-04
5  4.045435e-03 5.056794e-03 3.160496e-03 1.316873e-03 4.115229e-04 1.028807e-04
6  1.011359e-03 1.264198e-03 7.901240e-04 3.292183e-04 1.028807e-04 2.572018e-05
7+ 4.858139e-05 6.072674e-05 3.795421e-05 1.581426e-05 4.941955e-06 1.235489e-06
              6           7+
0  3.387020e-04 1.094781e-05
1  5.080530e-04 1.642171e-05
2  3.810397e-04 1.231628e-05
3  1.905199e-04 6.158140e-06
4  7.144495e-05 2.309303e-06
5  2.143349e-05 6.927908e-07
6  5.358371e-06 1.731977e-07
7+ 2.573935e-07 8.319685e-09

You could have also used a loop, as you originally suspected, but that's a more roundabout way to the same solution.

Upvotes: 1

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