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E[min (t, sigma y(i))] >= Sigma [q(i)/p(i)]
or in other words
E[min (t,i1+i2+i3+.........+in)*Pr(i1)Pr(i2)...........P r(in)]
i1,i2,i3,,,,,,,in goes from 1 to infinity
where y is a random variable following geometric distribution.
Is there a way to find a lower bound for t which would make the computation much easier.
Even for the case of i=2 i.e. y1+y2 the computation gets huge. Let alone for solving for a general n here. We have the input of
You are more likely to get a reply if you explain your notation (and provide more contextual information in general)Quote:
Originally Posted by himanshubahmani
CB
E[min(t, y1+y2+y3+...........+yn)] >= (q1/p1 + q2/p2 +........ qn/pn)
In this Problem q1,p1, .... qn,pn are all known
y1,y2........y3 are random variables following geometric distribution
Prob(y1=m)= p1x[(1-p1)^(m-1)]
For the case of y1 m can go from 1 to infinity, same from y2, y3.... and yn when we calculate the expectation value
e.g.
consider n=1 case
SIGMA [{min(t,i)} x Prob(i)]
when i goes from 1 to infinity
consider n=2 case
SIGMA SIGMA [{min(t, i1+i2)} x Prob(i1)Prob(i2) ]
where i1 and i2 can go from 1 to infinity
general case
SIGMA......... n times... Sigma [{min(t,i1+i2+i3.......+in)}x Prob(i1)xProb(i2)x........Prob(in)]
where i1,i2,i3..... in can go from 1 to infinity....