MHB Modifying frequency to eliminate 0

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The discussion centers on understanding how to modify a Poisson distribution to eliminate the case of zero profit in a scenario where customers arrive at a store at a Poisson rate. The user is confused about why the zero profit case is removed and how to identify such problems. It is clarified that the profit distribution is discrete and independent of the Poisson process for customer arrivals. The solution involves calculating the modified Poisson process for non-zero profits, which is essential for accurate probability calculations. Removing the probability of zero profit is crucial to focus on scenarios that yield positive earnings.
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Q: ppl arrive at Poisson rate of 0.5/min. Profit X is rand as follows:

Pr(X=0) = 0.7
Pr(X=1) = 0.1
Pr(X=2) = 0.1
Pr(X=3) = 0.1

Probability of 0 profit in 10min?
The start of the solution says to modify the Poisson to eliminate the case of 0 profit.
I don't understand why this is the case or how to identify this type of problem as such.
I don't see another way to reach the solution either...

Thanks!
 
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soran said:
The start of the solution says to modify the Poisson to eliminate the case of 0 profit.
I don't understand why this is the case or how to identify this type of problem as such.
I don't see another way to reach the solution either...

Thanks!

Hi soran,

Welcome to MHB! :)

You can scale a Poisson random variable in the following way:

If $X \sim \text{Poisson }(\lambda)$ where $\lambda$ is some rate per hour. Then $Y \sim \text{Poisson }(k \lambda)$ where $k \lambda$ is some rate per $k$ hours. Basically if you double the interval over which you are looking, you should have on average double the occurrences.

I would try thinking of this as a Poisson process with $\lambda=5$, which corresponds to 5 something per 10 minutes. If we call this process $Y$, what is the formula for $P(Y=0)$?

EDIT: I'm actually a little unsure of how $X$ corresponds to the Poisson process. Is this the whole problem? I ask because if $X \sim \text{Poisson }(\lambda)$ then $P(X=0)=0.6065307$.
 
Thanks it's good to be here :)

To clarify, the rate at which customers arrive at a store each min is Poisson~L but the profit that is earned from these customers is a discrete distribution X whose pdf is given.

I understand that 10*L=L* is also Poisson and L* is the param for 10min intervals. What I don't understand is the next part where the solution eliminates the probability of a profit of 0 as such:
Poisson process for non-zero profits is (10*L)[1-Pr(X=0)] = 1.5
Overall probability of zero profit is then = e^-1.5
The next question asks for the probability of making a profit of 2 in 10min.
The solution again removed the probability of 0 profit. Why is it so important to remove the probability of 0 both times?
 
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