Hi It appears to be the question of memory less propoerty but i'm simply not able to apply using simple probabilities q) If the probability is f(x) that a product fails the xth time it is being used .i.e. on the xth trial, then its failure rate at the xth trial is the probability that it will fail on the xth trial given that it has not failed in the x-1 trials is z(x) =f(x)/(1-F(x-1) where F(x) is the value of the corresponding distribution function of x. Show that if X is a geometric random variable, its failure rate is constant and equal to theta I have tried to proceed by using f(x)=geometric distribution but not able to show independence criteria. Can anyone show the steps pls.
thanks got it The trick was to get F(x). how did u get F(x) from f(x). I tried to integrate and getting a pretty bad expresion in terms of ln (1-theta). Also I see the formula for F(x) is not given in the actuarial table.
You get F(x) from the formula for a geometric progression. P(X=1) = theta P(X=2) = (1-theta)*theta P(X=2) = (1-theta)^2*theta etc P(X<=x) = theta + (1-theta)*theta + (1-theta)^2*theta + ..+ (1-theta)^x-1*theta Using sum of GP: F(x) = P(X<=x) = theta[1-(1-theta)^x]/1-(1-theta) which gives 1-(1-theta)^x.
