(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

I must show that the conditional density of probability [itex]P(t|\tau )dt[/itex] that a device fails between time t and t+dt given that it has no failed up to time [itex]\tau[/itex] is [itex]P(t|\tau )=\frac{P(t)}{S(\tau )}[/itex]; where [itex]P(t)dt[/itex] is the density of probability that the device will fail between between t and t+dt and [itex]S (\tau ) =\int _{\tau }^\infty P(u)du[/itex] is the probability that the system is reliable (i.e. did not fail) up to time [itex]\tau[/itex].

Let [itex]\gamma (t)=\lim _{\tau \to t}P(t|\tau )[/itex] then [itex]P(t)=\gamma (t) S(t)[/itex] so that [itex]\gamma (t )[/itex] can be thought as the rate of failure of the device.

Find [itex]P(t)[/itex] and [itex]S(t)[/itex] as functions of [itex]\gamma (t)[/itex].

Then, consider the cases when [itex]\gamma[/itex] is constant and where [itex]\gamma (t)=\delta (t-T)[/itex] for some positive T.

2. Relevant equations

P(A|B)=P(A intersection B)/P(B)

They forgot to mention that [itex]0 < \tau <t[/itex].

3. The attempt at a solution

I don't know whether the problem is extremely badly worded, confusing density of probability with probabilities or I don't understand anything.

Anyway, I've sought some help in Papoulis's book and here is my attempt.

[itex]\int _0^t P(t |\tau )dt = \frac{\int _0 ^t P(u)du - \int _0 ^\tau P(v) dv}{1-\int _0^\tau P(z)dz}[/itex]. Deriving with respect to t, I reach that [itex]P(t | \tau ) = \frac{P(t)}{S (\tau) }[/itex]. Which is the good result.

However I do not understand why the intersection of A and B in this case is [itex]\int _0 ^t P(u)du - \int _0 ^\tau P(v) dv[/itex] instead of [itex]\int _0 ^ \tau P(h)dh[/itex]. Can someone explain this to me?

For the next part, since [itex]S(t)=1-\int _0^t P(s)ds[/itex], then [itex]\dot S(t)=-P(t)[/itex]. Using the fact that [itex]\gamma (t)= \frac{P(t)}{S(t)}[/itex], I get that [itex]\gamma (t) S(t)=-\dot S(t)[/itex]. Solving that DE I reach that [itex]S(t)=\exp \left ( -\int _0^t \gamma (r)dr \right ) [/itex].

Hence [itex]P(t)=\gamma (t) \exp \left ( -\int _0^t \gamma (r)dr \right )[/itex].

The case [itex]\gamma[/itex] is a constant gives me [itex]S(t)=e^{-\gamma t}[/itex] and [itex]P(t)=\gamma e^{-\gamma t}[/itex].

The case [itex]\gamma (t)=\delta (t-T)[/itex] is much of a problem to me. It gives me [itex]S(t)=e^{-1}[/itex] if [itex]0 \leq T \leq t[/itex] and [itex]S(t)=1[/itex] if [itex]T>t[/itex]. But by intuition I'd have expected [itex]S(t)=0[/itex] instead of [itex]e^{-1}[/itex] for when [itex]0 \leq T \leq t[/itex]. That's one huge of a problem.

Second huge problem, [itex]P(t)=\delta (t-T) \exp \left ( -\int _0^t \delta (r-T) dr \right ) [/itex] which, to me, does not makes sense when not integrated. I mean I can't give any numerical value to this. I don't know how to deal with this.

Any help will be appreciated. Thank you!

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# Homework Help: Probabilities: System reliability

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