Question

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?

Was it helpful?

Solution

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.

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