Search Results

Here is some intuitive viewpoint different from simply adding the two exponential distributions: The point 0.5 was fixed, but we can also consider it as a randomly chosen point. In that case we may wo …

You can use the following approaches: The waiting time for a single event is exponential. The waiting time untill $k$ events is the sum of $k$ independent exponential distributed variables and is Erl …

You used the Poisson approximation of the Bernoulli distribution. For large values of $n$ probability close to zero, you get that the Bernoulli distribution is approximately equal to a Poisson distrib …

TL;DR Processes with a probability density of events in time or space that is constant (and independent from what happened before) have an exponential waiting time distribution, and from this you can …

$X$ will be distributed as a Skellam distribution. Intuition You can view it intuitively as following. For a Poisson process on some piece of length $L$, you randomly designate each event as $Y_i = +1 …

Computational Question 5 is similarly based on different reason It may often help to play a bit with some simple computations to get an idea how these arrivals look like. Below I am using the method …

Interpretation 2: No red and blue balls within the same 1m x 1m square of a grid In this case, the problem is similar to an urn model: you have $100^2$ urns in which you place red and blue balls, an …

Interpretation 1: No red and blue balls within any possible square of 1m x 1m #### plotting situation ### drawing boxes around the blue points box <- function(x, y , col = "blue") { polygon(c(x,x …

The negative binomial can be treated like Poisson, but it is ambiguous how to treat it. It will depend on the underlying process that causes the overdispersion. This can occur in different ways. Below …

Not necessarily a Poisson process The answer is already sort of given by WHuber in the comments. You need more (restricting) assumptions before the exponential waiting time is to be considered a Poiss …