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study probability problem that states the lengths of a classroom lecture follows an exponential distribution with mean $\frac{1}{\lambda}$, and iid so $f_X(x) = \lambda e^{(-\lambda*x)}$.

A student wanting to study the classroom lecture lengths wants to determine $\lambda$ and make an inference on this parameter. So he takes a random sample, but suspects that the randomly chosen observation times of the classroom lectures are all longer than the expected/average classroom times.

Assuming that his sample of n observations T1,...,Tn, were sampled such that all were longer than the average time, find the conditional distribution/density of one observation.

So my first attempt to apply the memoryless property, since each observation is at least as long as the expected classroom time ($1/\lambda$), then $T_i > s$ for all i, and $s=\frac{1}{\lambda}$. so I have $P(T>t | T> ET)=P(T>t|T>\frac{1}{\lambda})= \frac{P(T>t)}{P(T>1/\lambda)} = e^{-\lambda*t+\lambda(\frac{1}{\lambda})}=e^{-\lambda*t+1}$

So the CDF $P(T\leq t) = 1-e^{-\lambda*t+1}$ and from here I can derive the PDF.

This seems correct to me, but the other way is to condition on the event that $E=\{T>1/\lambda\}$ (but got stuck here). Using Bayesian theorem $P(T=t | T>\frac{1}{\lambda}) = \frac{P(T>\frac{1}{\lambda} | T=t)P(T=t)}{P(T>1/\lambda)}$=$\frac{P(T>\frac{1}{\lambda}+t)f_T(t)}{P(T>1/\lambda)}$

is this on the right track? any hint as to what i'm missing is appreciated.

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    $\begingroup$ I might be wrong, but I think you are interpreting the problem incorrectly. There are $n$ observations $T_1, \ldots, T_n$, and you are considering a separate new observation $T$. They want you to compute the distribution of the new observation $T$ given $\{T_1 > 1/\lambda\} \cap \cdots \cap \{T_n > 1/\lambda\}$. $\endgroup$
    – angryavian
    May 20, 2021 at 23:16

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My interpretation is that the distribution would be $f(x|x>\frac 1\lambda)=\lambda e^{-\lambda(x-\frac 1 \lambda)}$, as due to the memorylessness property the pdf is shifted to the right. We're given that the observations are all at least $\frac 1 \lambda$ so the pdf should "start" at $\frac 1 \lambda$.

The other way to do it is by "reweighting" the density: $$\frac{\lambda e^{-\lambda x}}{P(X>\frac 1 \lambda)}=\frac{\lambda e^{-\lambda x}}{e^{-\lambda(\frac 1 \lambda)}}=\lambda e^{-\lambda x+1}, x>\frac 1 \lambda$$

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    $\begingroup$ should it be $\lambda e^{-\lambda(x-1/\lambda)}$ ? I think this is right, we essentially need to find $f_X(x-1/\lambda)$ ? $\endgroup$ May 20, 2021 at 23:43
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    $\begingroup$ @sophie-germain definitely missed the minus sign there, thanks $\endgroup$
    – Jellyfish
    May 20, 2021 at 23:46
  • $\begingroup$ Yup. And I am not sure how to apply the bayes rule here but I do know that you can do something to the density to get $f(x|x>1/\lambda)$, see the post. $\endgroup$
    – Jellyfish
    May 21, 2021 at 0:02

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