Re: how to model a probability distribution in this case?
- From: challengerlee <lihui0706@xxxxxxxxx>
- Date: Mon, 3 Dec 2007 01:28:17 -0800 (PST)
On 12月3日, 下午4时07分, Ray Vickson <RGVick...@xxxxxxxx> wrote:
On Dec 2, 11:08 pm, challengerlee <lihui0...@xxxxxxxxx> wrote:
Assume there are ten bulbs, whose lifetime follows the exponential
distribution with parameter b_{i} (1<=i<=10).
Then here is the problem, what is each bulb's probability of being the
first one to be broken in the following period?
My intuition tells me, since the mean lifetime of each bulb is 1/
b_{i}, then the corresponding probability is P(i)=1-(1/a_{i})/(1/
a_{1}+...+1/a{10}), but i cannot figure the reason.
What are the a_{i}? You started with parameters b_{i}.
So can anybody
tell me the reason or a better way to get this probability?
Look first at the case of n = 2 bulbs, with rate parameters b1 and b2
and independent random exponential lifetimes X1 and X2. We have: (i)
the time to first failure is exponential with rate b1+b2 (because the
first failure occurs at time X = min(X1,X2); (ii) the probability that
bulb 1 fails first is b1/(b1+b2), because P{X1<X2} = integral(b1*exp(-
b1*x)*P{X2>X1|X1=x} dx, and P{X2>X1|X1=x} = P{X2>x|X1=x} = P{X2>x},
since X1 and X2 are independent. Do the integral to get the result.
Next, try it with three bulbs; you should find P{X1 is smallest} = b1/
(b1+b2+b3), because min(X2,X3) = exponential(b2+b3) and P{X1 is
smallest} = P{X1 < min(X2,X3)}. All this is standard material, found
in any halfway decent textbook, or on web pages about exponential
distributions.
R.G. Vickson
thanks!!! i get it.
.
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