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October 20

floor( exponential random variable ) = geometric distribution

How can I prove that if one takes the floor function of an exponential random variable it yields a geometric distribution? I've read in many places that it is really true, but I cannot find a proof, no matter how much I search. Thanks. --Belchman (talk) 20:51, 20 October 2010 (UTC)[reply]

Interpret the variable as time (in, say, minutes): since the distribution is memoryless, after a minute has passed, the chances of the event occurring in the second minute are the same as it had at the start of happening in the first minute. That the distribution of floors is geometric with the parameter equal to that minute's probability follows immediately. --Tardis (talk) 21:40, 20 October 2010 (UTC)[reply]

You can also use the direct approach: Let

X\sim \exp(\lambda ),Y=\lfloor X\rfloor \;\!

. Then

\mathrm {Pr} (Y=k)=\mathrm {Pr} (k\leq X<k+1)=\int _{k}^{k+1}\lambda e^{-\lambda x}\ dx=e^{-k\lambda }(1-e^{-\lambda }).

If you take

p=1-e^{-\lambda }

then this is

(1-p)^{k}p

which is a geometric distribution. -- Meni Rosenfeld (talk) 09:13, 21 October 2010 (UTC)[reply]

Thank you. Brilliant answers. --Belchman (talk) 18:47, 21 October 2010 (UTC)[reply]