interview question

dreamonstreet · August 29th, 2008, 2:47 am

x1 and x2 are i.i.d. random variables on [0,\infinity) with pdf. f(x)=sqrt(2/pi)*e^(-x^2/2). The question is to findE[max{x1,x2}/min{x1,x2}], i.e., the expected ratio between the big number and the small number.

quantyst · August 29th, 2008, 9:50 am

Under Modification! See My Next Posts.

quantyst · August 29th, 2008, 9:54 am

QuoteOriginally posted by: dreamonstreetx1 and x2 are i.i.d. random variables on [0,\infinity) with pdf. f(x)=sqrt(2/pi)*e^(-x^2/2). The question is to findE[max{x1,x2}/min{x1,x2}], i.e., the expected ratio between the big number and the small number.Maybe it would be more interesting to compute E[min{x1,x2}/max{x1,x2}].

amit7ul · September 2nd, 2008, 12:21 pm

or equivalently E ( max ( x1/x2, x2/x1 ) ) and x1/x2 for x1,x2~N is a well known distribution is it.. would that help

NicolasQuant · September 2nd, 2008, 1:13 pm

+Infinity?

quantyst · September 2nd, 2008, 4:52 pm

Here's a commando approach:We know max{x1,x2}/min{x1,x2} >= (x2 / x1); hence E[max{x1,x2}/min{x1,x2}] >= E[x2 / x1]=Infinity.

NicolasQuant · September 2nd, 2008, 9:32 pm

You were quicker than me quantyst.I had to write down the integral just to figure out it was... indefinite