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My question pertains to the example shown in: https://mathtopics.wordpress.com/2012/12/03/going-bankrupt/

I am having trouble understanding the difference between $P(Y)$ and $p_Y$. The article states that the former is "What are your chances of getting “Yes”?" and the latter is "probability that the robot answers “Yes”. What is the difference between these two? They seem to mean the same thing to me.

My next question pertains to the summation term for P(Y). I believe this summation came from the law of total probability, but the variable we're dealing with here is continuous, so why did they start it out as a summation rather than an integral?

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Note that the robot picks up a random probability each time you ask, namely $p_Y$, and also says 'Yes' with that probability. Here, $P(Y)$ is the final probability that you get a 'Yes' from the robot and $p_Y$ is a random variable. So, you have a conditional probability, $P(Y|p_Y)=p_Y$, and to calculate the overall probability of getting a 'Yes', you need to apply total probability law:

$$P(Y)=\int_0^1 P(Y|p_Y=p_y)f_{p_Y}(p_y)dp_y=\int_0^1 xf(x)dx$$

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  • $\begingroup$ I think that makes sense. Just a few more questions (1) Is it correct to assume that the $f$ pdf they provided is a uniform distribution, hence the limits of integration are from 0 to 1? (2) What was the point of them starting out with a summation as opposed to an integral? Though I suppose if the pdf is a uniform distribution then you can easily split it up into rectangles to find the area using numerical integration (which is what I think they tried to show). (3) Is it correct to say that $P(Y)$ is the probability over a large number of draws and $p_Y$ is a specific draw? $\endgroup$ Mar 30, 2020 at 13:42
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    $\begingroup$ (1)It doesn't have to be uniform. If it was $P(Y)$ would be $0.5$. The limits are from $0$ to $1$ because the RV $p_Y$ is a probability. (2) Yes, the summation represents a discretised integral, and it's a rather confusing way to compared to Total Law of Prob I've used. (3) Kind of. P(Y) is the probability of getting Yes from the robot when you don't know anything about its current 'yes probability', so we report the average value of $p_Y$ (as if we asked the robot a lot of times) and $p_Y$ is the probability of getting yes, when we know the probability of 'Yes' of the robot, so specific draw. $\endgroup$
    – gunes
    Mar 30, 2020 at 13:54

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