- #1
rsq_a
- 107
- 1
I did this question, but I'm unsure of my reasons behind it. I was hoping someone here could go through the problem for me.
I got the answer [tex]1/\lambda - 1/(\lambda + \mu)[/tex]. I did so by integrating,
[tex]\int_0^\infty P(\text{one event from } \lambda \text{ in }(0, t]) \times P(\text{zero event from } \mu \text{ in }(0, t]) \ dt [/tex]
Except I didn't have any good reason for integrating the whole thing except for the idea that I want to add up all the probabilities. Is this the way it's supposed to be done?
Consider the sum of two independent Poisson processes of rates [itex]\lambda[/itex] and [itex]\mu[/itex]. Find the probability that the first arrival of the combined [itex](\lambda + \mu)[/itex] process comes from the process of rate [itex]\lambda[/itex]
I got the answer [tex]1/\lambda - 1/(\lambda + \mu)[/tex]. I did so by integrating,
[tex]\int_0^\infty P(\text{one event from } \lambda \text{ in }(0, t]) \times P(\text{zero event from } \mu \text{ in }(0, t]) \ dt [/tex]
Except I didn't have any good reason for integrating the whole thing except for the idea that I want to add up all the probabilities. Is this the way it's supposed to be done?