- #1
Ocifer
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Hello, hopefully this is the right place. This is a homework question, so it should definitely be in this forum, but I wasn't sure which sub-forum to put this rather elementary stats question.
In my introductory mathematical statistics class, we've been given the following word problem having to do with Poisson processes.
We are to consider a football game, where each team's score follows its own Poisson process. So there are two teams, team A and team B and, each team's final score has its own lambda parameter:
Team A scores \lambda_A goals per game.
Team B scores \lambda_B goals per game.
What is the probability of team B winning over team A, with a final score of 3-2.
My intuition (which I would like to confirm/have critiqued) is that since the processes are independent of each other I should find the following.
Let X_A be the final score of team A
Let X_B be the final score of team B
<br /> Pr(X_A = 2) = \lambda_A ^ 2 e^{- \lambda_A } / 2!<br />
<br /> Pr(X_B = 2) = \lambda_B ^ 3 e^{- \lambda_B } / 3!<br />
And then the probability that team B wins with a score of 3 to 2 is simply the product of the two probabilities above, namely:
<br /> Pr(X_A = 2 \cap X_B = 3) = Pr(X_A = 2) \cdot Pr(X_B = 3)<br />
Is this the correct approach?
Homework Statement
In my introductory mathematical statistics class, we've been given the following word problem having to do with Poisson processes.
We are to consider a football game, where each team's score follows its own Poisson process. So there are two teams, team A and team B and, each team's final score has its own lambda parameter:
Team A scores \lambda_A goals per game.
Team B scores \lambda_B goals per game.
What is the probability of team B winning over team A, with a final score of 3-2.
Homework Equations
The Attempt at a Solution
My intuition (which I would like to confirm/have critiqued) is that since the processes are independent of each other I should find the following.
Let X_A be the final score of team A
Let X_B be the final score of team B
<br /> Pr(X_A = 2) = \lambda_A ^ 2 e^{- \lambda_A } / 2!<br />
<br /> Pr(X_B = 2) = \lambda_B ^ 3 e^{- \lambda_B } / 3!<br />
And then the probability that team B wins with a score of 3 to 2 is simply the product of the two probabilities above, namely:
<br /> Pr(X_A = 2 \cap X_B = 3) = Pr(X_A = 2) \cdot Pr(X_B = 3)<br />
Is this the correct approach?
