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I have a problem where I am working out the failure time of a system before time $t$. I have the probability of this happening:

$$ 1 − R(t) + R(t)(1 − R(t))^2(1 − R(t)^2) . $$

How can I work out the density function and mean of failure time from this

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    $\begingroup$ Your notation is unclear. What is R(t)? $\endgroup$
    – Adrian
    Jan 13, 2016 at 13:42

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The PDF

What you have is the cumulative distribution function, which is $\int_0^t{p(\tau) d\tau}$, if your function represents the probability of failure before time t. (Where $p(\tau)$ represents the pdf). We can view the cdf as accumulating the probability of failure for all time.

The probability density function can be found by taking the derivative of the above function. We can intepret this as determining the probability of failure at some time t. So the probability of a failure occuring between any two times is $\int_{t_1}^{t_2}{p(\tau)d\tau}$

$ cdf(t) = \int_0^t{p(\tau) d\tau}$

$\frac{d}{dt} cdf(t) = p(t)$

In summary, the derivative of your wacky function of R should directly yield your pdf in time.

The Mean

Since p is a density, the mean $\mu$ can be computed as

$\mu = \int_{-\infty}^\infty{t*p(t)}$

Since for all t<0, p is 0, you only have to evaluate that integral from 0 to $\infty$

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