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Definition/Summary
The exponential distribution is a probability distribution that describes a machine that it equally likely to fail at any given time.
Equations
[tex] f(t) = e^{-\lambda t} \lambda [/tex]
Extended explanation
A machine is equally likely to fail at any given time. For any [itex]t[/itex], the probability of failure in the interval [itex](t, t + dt)[/itex] is [itex]\lambda dt[/itex]. So the probability that it doesn't fail in that interval must be [itex]1 - \lambda dt[/itex]
Let us calculate the probability that it doesn't fail in the interval [itex](0,t)[/itex]. Divide [itex]t[/itex] into [itex]n[/itex] equal parts. Each part then has size [itex]\frac{t}{n}[/itex]. The probabilities that it doesn't fail in the intervals [itex](0,\frac{t}{n})[/itex], [itex](0,2\frac{t}{n})[/itex], ... are
[tex] 1 - \lambda \frac{t}{n} [/tex]
[tex] (1 - \lambda \frac{t}{n})^2 , [/tex]
... respectively. Therefore we find that the probability that it doesn't fail in the interval [itex](0,t)[/itex] is approximately
[tex] (1 - \lambda \frac{t}{n})^n .[/tex]
The exact answer is the limit of the above expression as [itex] n \rightarrow \infty[/itex], i.e.
[tex]e^{-\lambda t}.[/tex]
We can now use this to find the probability density function [itex]f(t)[/itex] that it fails for the first time in the interval [itex](t,t+dt)[/itex]. Clearly, this is the probability that it doesn't fail in the interval [itex](0,t)[/itex] times the probability that it fails in the interval [itex](t,t+dt)[/itex].
[tex]f(t) = e^{-\lambda t} \lambda.[/tex]
* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
The exponential distribution is a probability distribution that describes a machine that it equally likely to fail at any given time.
Equations
[tex] f(t) = e^{-\lambda t} \lambda [/tex]
Extended explanation
A machine is equally likely to fail at any given time. For any [itex]t[/itex], the probability of failure in the interval [itex](t, t + dt)[/itex] is [itex]\lambda dt[/itex]. So the probability that it doesn't fail in that interval must be [itex]1 - \lambda dt[/itex]
Let us calculate the probability that it doesn't fail in the interval [itex](0,t)[/itex]. Divide [itex]t[/itex] into [itex]n[/itex] equal parts. Each part then has size [itex]\frac{t}{n}[/itex]. The probabilities that it doesn't fail in the intervals [itex](0,\frac{t}{n})[/itex], [itex](0,2\frac{t}{n})[/itex], ... are
[tex] 1 - \lambda \frac{t}{n} [/tex]
[tex] (1 - \lambda \frac{t}{n})^2 , [/tex]
... respectively. Therefore we find that the probability that it doesn't fail in the interval [itex](0,t)[/itex] is approximately
[tex] (1 - \lambda \frac{t}{n})^n .[/tex]
The exact answer is the limit of the above expression as [itex] n \rightarrow \infty[/itex], i.e.
[tex]e^{-\lambda t}.[/tex]
We can now use this to find the probability density function [itex]f(t)[/itex] that it fails for the first time in the interval [itex](t,t+dt)[/itex]. Clearly, this is the probability that it doesn't fail in the interval [itex](0,t)[/itex] times the probability that it fails in the interval [itex](t,t+dt)[/itex].
[tex]f(t) = e^{-\lambda t} \lambda.[/tex]
* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!