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With the information you have now you don't really need Bayes. By the basic definition of conditional probability $$ P(B \mid M, S) = \frac{P(M, S, B)}{P(M, S)} = \frac{P(\text{Male, smiling, bought beer})}{P(\text{Male, smiling})} = \frac{10/60}{15/60}= \frac{2}{3} $$


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"If B happens given A, didn't A and B happen at the same time?" Yes that is true (or it could be true--it could also be considering sequential events--not necessarily simultaneous ones). However, A and B happening together might be "unlikely" (really it just means they might not always happen together). So consider that you have three outcomes and exactly ...


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The difference is that in one case, you already know B has happened. This is not subject to chance anymore, it's a fact. So you compute the probability of A happening, knowing for a fact that B happened For the probability in the intersection, on the other hand, you don't know that B has happened and have to calculate the probability that both A and B ...


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Your current data is insufficient to answer, and this is why you run into errors. You have a discrete distribution (Beer, noBeer) | (Male, Female) | (Smile, noSmile), so we can think of the joint probability distribution as being a 2*2*2 array (or a Tensor in math-speak) Right now, the data you are giving specifies only the joint distribution of Beer and ...


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The basic idea is to chop up the convex polytope into a bunch of simplices, then to pick one simplex at random (fairly) and then to sample within the simplex. The hardest part is determining the vertices of the polytope (which you need to triangulate it), since you have a facet-based description (inequalities) rather than a vertex-based description, and the ...


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The answer depends slightly on your wording. You say that you get heads every time. Now if you had not decided in advance that you would look for heads every time and you asked the question how many times would you have to see the same side of the coin show up to conclude that it is biased at the 99% level, then assuming that you have never performed any ...


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$$p(A)=1-\frac{1}{2^n}$$ so you need to $$p>99%\\ $$ or $$p>\frac{99}{100}\rightarrow 1-\frac{1}{2^n}>\frac{99}{100}\\\frac{1}{2^n}<\frac{1}{100} \\2^n>100\\n>7$$


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You need to apply Baye's rule several times: $$P\{G|A,M\}=\frac{P\{G,A,M\}}{P\{A,M\}}=\frac{P\{A|G,M\}\cdot P\{G,M\}}{P\{A,M\}}=\frac{P\{A|G,M\}\cdot P\{G|M\}\cdot P\{M\}}{P\{A|M\}\cdot P\{M\}}$$


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Try $X$ with a uniform distribution on $[0,\theta]$ where $\theta$ has a Pareto distribution with density $\dfrac{\alpha m^\alpha}{\theta^{\alpha+1}}$ when $\theta \gt m$ (and both $m \gt 0$ and $\alpha \gt 0$). The conditional expectation would then be affected by the maximum of the observations rather than some linear combination of them all.


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To put it a bit more formally... Let $p_r$ be the prior probability for rain ($p_n=1-p_r$). Then the probability of rain given 3 "yes" replies $(y,y,y)$ $$P\{\text{rain}|y,y,y\}=\frac{P\{\text{rain}\bigcap (y,y,y)\}}{P\{y,y,y\}}$$ $$=\frac{P\{y,y,y|\text{rain}\}\cdot p_r}{P\{y,y,y|\text{rain}\}\cdot p_r+P\{y,y,y|\text{no rain}\}\cdot p_n}$$ Next, if we ...


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You are completely right about Bayesian analysis. It is indeed correct to say that the probability is 8/27 (not 8/9) that they all tell the truth, prior to the experiment. However, post the experiment, you have three answers. If in the area it is unlikely to rain (for interest in the desert), the three 'yes' answers are proof to two things: a) it might ...


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For a Poisson likelihood, the gamma distribution is a conjugate prior, so the posterior is also gamma, as you have found. This lets you calculate the posterior in two different ways: one is to compute $f_{\Lambda}(\lambda \mid x)$ for a single observation, recognize this as gamma with updated hyperparameters, and then use an inductive argument to get ...


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It's basically Bayes Rule. Recall the Bayes Rule formula: $P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{P(B|A)P(A)}{\int P(B|A)P(A)dA}$ Basically, we can think of $h$ as $P(B|A)P(A)$ But, what we're really interested in is $P(A|B)$ (the probability of being type A conditional on information B) and in order to get in we need to divide by $\int P(B|A)P(A)dA$ Edit: ...


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First, if the probability of heads on a single toss is $Y$, then $Y \in [0,1]$, hence $$1 = \int_{y=0}^1 f_Y(y) \, dy = \int_{x=0}^1 ky^2 \, dy = \frac{k}{3},$$ so $k = 3$ and the density of $Y$ is $$f_Y(y) = 3y^2, \quad 0 \le y \le 1.$$ This characterizes the prior distribution of the probability of heads, and is actually a special case of a Beta ...


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If you believe the three sources to be equally reliable and that they are the only ones to be considered, then maybe it is reasonable to start with $P(A) = P(B) = P(C) = 1/3.$ Predictions are $P(R|A) = 0,\,$ $P(R|B) = 1,\,$ and $P(R|C) = 1/2.$ Then you might use the law of total probability to get $P(R),\,$ and carry an umbrella tomorrow.


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In statistical computations that find the sample size or degrees of freedom, or that estimate integer parameters, do not have to result in integer values. For sample sizes, it is customary to round up to the next higher integer. For degrees of freedom in using printed t and F tables, it is customary to round down to the next lower integer. (Both ...


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First, The representation $p(x)=\int{f(x|\alpha,\beta)g(\alpha)h(\beta)}$ can span the entire $[l,u]$ segement if $g()$ and $h()$ are allowed to be negative. This can be seen by extending polynomial approximation. However, if $h,g \geq 0$, the answer is simply no. $p(x)\equiv 1$ cannot be obtained


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Let's take the case of the beta distribution with parameters $\alpha,\beta>0$: $$ f(x;\alpha,\beta)=\frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha,\beta)} \quad\text{where}\quad B(\alpha,\beta)=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}. $$ The question is if a continuous probability distribution $p(x)$ on $[0,1]$ can always be expressed in ...



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