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rasmodius
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How does an observation condition the next one, if the numbers are exp. distributued with uknown average?
@BruceET, I'm sorry but there's not more context. It is neither a textbook nor a practical problem. I just made up the problem and was curious about the solution.
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How does an observation condition the next one, if the numbers are exp. distributued with uknown average?
@BruceET, if I asked just for you an estimation of $x_2$, you can't say anything of course. However, imagine that you had a prior observation with $x_1=112$, and that you knew that the values are exponentially distributed. In these circumstances, you know that the value $x_2=10^5$ is much less probable than $x_2=80$ because the value $x_1=112$ is likely close to the unknown average of the exponential distribution from where all the values are being drawn. Therefore, knowing $x_1$ changes the probability for $x_2$, understanding the concept of probability as degree of confidence (bayesian way).
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How does an observation condition the next one, if the numbers are exp. distributued with uknown average?
@BruceET, thanks for your answer. However, we are looking for the estimation or, ideally, the probability distribution of $x_2$, not of $\lambda$.
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How does an observation condition the next one, if the numbers are exp. distributued with uknown average?
Yes, they are assumed, independent. Regarding the prior, I would use Jeffrey's scale prior between 0 and infinity since is the most uninformative one in this case.
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