June 22 Information

I have a large database from which the records are deleted at the rate of 1/10⁵ every seven years. If I pick five records at random, what is the expectation value for the time at which I will have only three records left? How long to go from three to two? Spinning Spark 23:51, 22 June 2012 (UTC) reply

See our article on exponential decay. Bo Jacoby ( talk) 11:53, 23 June 2012 (UTC). reply

When there are n records, and each decays at a rate of ${\displaystyle 1/(7\cdot 10^{5})}$, together they decay at a rate of ${\displaystyle n/(7\cdot 10^{5}\mathrm {yr} )}$, thus it will take on average ${\displaystyle (7\cdot 10^{5})/n}$ years to go to ${\displaystyle n-1}$. Thus from 5 to 3 is 315,000 years and from 3 to 2 is 350,000 233,333.3. -- Meni Rosenfeld ( talk) 16:58, 24 June 2012 (UTC) reply

Thanks Meni, that's very helpful. Spinning Spark 23:54, 24 June 2012 (UTC) reply

Actually, shouldn't 3 to 2 (7/3) be 233,333 years? Spinning Spark 00:01, 25 June 2012 (UTC) reply

Right, fixed. -- Meni Rosenfeld ( talk) 09:27, 27 June 2012 (UTC) reply