I am trying to solve this question:

Let $f_T (t\mid \lambda) = \lambda e^{−\lambda t}$, $t ≥ 0$

So $T$ is exponentially distributed when conditioned on $\lambda$. Assume that we have a prior for $\lambda$ that is continuous uniform between $a$ and $b$.

(a) Given the event $E = \{T > \tau\}$, what can we say about $\lambda$? In other words, find $f(\lambda\mid E)$. Recall that Bayes rule allows for mixing probabilities and densities.

In particular: $$f(\lambda\mid E) = \frac {P(E\mid\lambda)f(\lambda)}{P(E)}.$$

I am a bit confused as to how I would go about solving this problem. I understand that $T$ is a random variable which has an exponential distribution given by $F_T$. I'm a little confused as to what "prior for $\lambda$ is continuous uniform between $a$ and $b$ means" my interpretation is that $\tau$ is the prior and is itself a random variable which is uniformly distributed.

Does this then imply that event $E$ is also uniform?

I also can't figure out what the distribution of $f(\lambda)$ is (is it just constant?).

The main problem I have is understanding what $F_T$ actually is. This question is given in the context of a random event happening within a time interval. $\lambda$ I believe is the base probability of an event happening (in which case its distribution would be constant) and $t$ is the amount of time before an event (in which case it would make sense that it is uniform, since time is measured in uniformly distributed equal intervals).

Then, $F_T$ I interpret as, given that a random event has happened, the amount of time that has passed till that event, is exponentially distributed. Therefore, for example, large values of $t|\lambda$ would be unlikely to occur, which makes sense since the probability that the event will happen will quickly increase with time.

I'm having the most difficult understanding event E, what does $T>\tau$ mean? And what does $f(\lambda \mid E)$ represent?


First off, an event is a subset of the sample space; it doesn't make sense to talk about a probability distribution for $E$.

As for what "prior" means, $T$ is an exponential random variable with unknown parameter $\lambda$. The problem is saying that $\lambda$ itself is a random variable, with $U(a,b)$ distribution. So the probability density function of is $\lambda$ is $f(x) = \mathsf 1_{(a,b)}(x)$ (with $\mathsf 1$ denoting the indicator function). Now, given a value of $\lambda$, we have

\begin{align} \mathbb P(E\mid\lambda) &= \mathbb P(T > \tau\mid \lambda)\\ &= \int_\tau^\infty \lambda e^{-\lambda t}\ \mathsf dt\\ &= e^{-\lambda\tau}. \end{align} Note that $\tau$ is just an arbitrary positive real number. Further, \begin{align} \mathbb P(E) &= \int_{\mathbb R} \mathbb P(T>\tau\mid \lambda)f(\lambda)\ \mathsf d\lambda\\ &= \int_{\mathbb R} \int_{\tau}^\infty f(t\mid\lambda)f(\lambda)\ \mathsf dt\ \mathsf d\lambda\\ &= \int_{\mathbb R} \int_{\tau}^\infty \lambda e^{-\lambda t}\mathsf 1_{(a,b)}(\lambda)\ \mathsf dt \ \mathsf d\lambda\\ &= \int_a^b e^{-\lambda\tau}\ \mathsf d\lambda\\ &= \frac1\tau (e^{-\lambda a}-e^{-\lambda b}). \end{align} Hence $$ f(\lambda\mid E) = \left(\frac{\tau e^{-\lambda\tau}}{e^{-\lambda a}-e^{-\lambda b}}\right)\mathsf 1_{(a,b)}(\lambda). $$

  • $\begingroup$ While this is definitely an answer, I was really looking for an intuitive explanation (in regards to the context of time for instance). I also don't understand how you conclude that $f(\lambda)$ is a continuous, uniform random variable from the fact that its prior is uniform and continuous. Also, for $P(E)$, how did you get the integral of $f(t|\tau)$? Isn't $P(T>\tau)$ solved above to be $1-e^{-\lambda t}? $\endgroup$ – Dider Oct 3 '15 at 20:28
  • $\begingroup$ Also, for the definite integral in the last step (from a to b) how did you get that answer? I tried integrating on paper and using Wolfram Alpha and both returned: $([e^(-a\tau) - e^(-b\tau)] \ \tau$. If you are integrating with respect to $\lambda$ shouldn't you substitute a and be for lambda? $\endgroup$ – Dider Oct 3 '15 at 20:47
  • $\begingroup$ Ah, I felt like something was off. Thanks, I'll correct the answer when I have time later. $\endgroup$ – Math1000 Oct 3 '15 at 22:29
  • $\begingroup$ Actually the first integral also seems to be in error, since the $\lambda$ is a constant, it can just be factored out of the integral and you get just $-e^{-\lambda t}$ (for the indefinite integral) $\endgroup$ – Dider Oct 4 '15 at 16:34
  • $\begingroup$ Yes, indeed it is. Thanks once again! $\endgroup$ – Math1000 Oct 4 '15 at 17:33

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