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Given some quantity $z$ which takes on two values, either $1$ or $2$ with probabillity $P(1) = 0.3$. Consider that there is some other measurement $y$ which takes on values $U$ or $V$ where we have that if we obtain measurement $V$ then we know that the probability of getting $z=1$ is $0.5$. Assuming we made a measurement and obtained $V$, to find the likelihood function for the measurement, I considered using Bayes Theorem in the following way: $$P(V|1) = \frac{P(1|V)P(V)}{P(1)} = \frac{(0.5)P(V)}{(0.3)}$$ Hence in order to know the likelihood function it seems I have to have knowledge of $P(V)$? Is this correct or can I make more progress given the knowledge that I outlined?

Thanks for any help.

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You're right, there's not enough information here. To see this, you can come up with two different values of $P(V)$ that are entirely consistent with the information you have given. Then, to complete the picture, you can calculate $P(1 \mid U)$ using, $$ P(1) = P(1 \mid U) P(U) + P(1 \mid V) P(V), $$ and substituting $P(U) = 1 - P(V).$ Try this for $P(V) = 0$ and $P(V) = 0.5.$ You find that there are no contradictions with your initial information, incurred from completing the picture from both cases. Thus $P(V)$ is undetermined from your information.

(Notice that you cannot use the case $P(V) = 1.$ See why that is. In fact, there is an upper limit to the value $P(V)$ can take. This has to do with the value that $P(1 \mid U)$ must take, given the value of $P(Z).$)

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  • $\begingroup$ Okay I see thanks, but would you say that our uncertainty in $z$ has increased or decreased after our measurement? It seems to me that our certainty in $z =1$ has increased we could say? $\endgroup$
    – Moses
    Jun 1, 2017 at 15:44
  • $\begingroup$ I'm not sure what you mean by "uncertainty". If you mean the variance of the distribution over $z,$ then the variance decreases if we observe $y=U.$ However, counter-intuitively, the variance would actually increase if we observed $y=V.$ This is because, for a Bernoulli distribution, a probability of $0.5$ is what maximizes the variance. Learning that $y=V$ is like one of those confounding revelations that makes you realize you know less about the system than you thought. If you want, I can explain why in my answer. $\endgroup$
    – Bridgeburners
    Jun 1, 2017 at 16:48
  • $\begingroup$ Just to confirm. The variance is defined as $$\sigma^2 = \sum(z_i-\mu)^2p(z_i).$$For the case where we don't measure $V$ I get $\sigma^2 = 0.21$ and for the case where we measure $V$ I get $\sigma^2 = 0.25$. $\endgroup$
    – Moses
    Jun 1, 2017 at 18:19

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