elemprob-fall2010-page36

# elemprob-fall2010-page36 - this constant over the set where...

This preview shows page 1. Sign up to view the full content.

One can conclude from this that f X,Y ( x,y ) = f X ( x ) f Y ( y ) , or again the joint density factors. Going the other way, one can also see that if the joint density factors, then one has independence. An example. Suppose one has a ﬂoor made out of wood planks and one drops a needle onto it. What is the probability the needle crosses one of the cracks? Suppose the needle is of length L and the wood planks are D across. Answer. Let X be the distance from the midpoint of the needle to the nearest crack and let Θ be the angle the needle makes with the vertical. Then X and Θ will be independent. X is uniform on [0 ,D/ 2] and Θ is uniform on [0 ,π/ 2]. A little geometry shows that the needle will cross a crack if L/ 2 > X/ cos Θ. We have f X, Θ = 4 πD and so we have to integrate
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: this constant over the set where X < L cos Θ / 2 and 0 ≤ Θ ≤ π/ 2 and ≤ X ≤ D/ 2. The integral is Z π/ 2 Z L cos θ/ 2 4 πD dxdθ = 2 L πD . If X and Y are independent, then P ( X + Y ≤ a ) = Z Z { x + y ≤ a } f X,Y ( x,y ) dxdy = Z Z { x + y ≤ a } f X ( x ) f Y ( y ) dxdy = Z ∞-∞ Z a-y-∞ f X ( x ) f Y ( y ) dxdy = Z F X ( a-y ) f Y ( y ) dy. Diﬀerentiating with respect to a , we have f X + Y ( a ) = Z f X ( a-y ) f Y ( y ) dy. There are a number of cases where this is interesting. (1) If X is a gamma with parameters s and λ and Y is a gamma with parameters t and λ , then straightforward integration shows that X + Y is a 36...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online