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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: <pre><code>name <- c("0","1","2","3","4","5","6","7+") lambda1 <- 1.5 lambda2 <- 1.25 </code></pre> We can get the marginal probabilities for 0-6 by using <code>dpois</code>, and the marginal probability for 7+ using <code>ppois</code> and <code>lower.tail=FALSE</code>: <pre><code>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)) </code></pre> 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: <pre><code>p.xy <- outer(p.x, p.y) rownames(p.xy) <- colnames(p.xy) <- name </code></pre> Now you're done: <pre><code> 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 </code></pre> 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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