Tag Info

New answers tagged

0

The middle step comes from Bayes' rule: $$ \pi(\theta | x) = \frac{\pi(x | \theta) \pi(\theta)}{\pi(x)} $$ where the denominator is an integral over $\theta$.


0

Kadane's book Principles of Uncertainty derives Bayesian probability from first principles and does use measure theory (in section 4.9). It is based on DeFinetti's framework of coherence instead of Cox's axioms.


2

The question's definitions of "false positive rate" and "false negative rate" look wrong and misleading to me. The false positive rate should be the proportion of people actually without cancer who give a false positive result on the test. Similarly the false negative rate should be the proportion of people actually with cancer who give a false negative ...


0

There's no problem with the answer. Bayes formula: $$ p(t|\pi^*) \propto p(\pi^*|t)p(t). $$ Since $p(\pi^*|t)$ is $1$ only when $t\in [-1, \pi^*]$ and $0$ otherwise, you get the resulting posterior. The posterior is clearly not uniform on $[-1,1]$; you're misquoting the result on conjugate distributions. Player $B$ knows player $A$ signals only when $A$'s ...


3

Consider the Theorem of Bayes for two competing events $A, B$ (with "$c$" for $complement$): $P(B|A) = \dfrac{P(A|B) \cdot P(B)}{P(A|B) \cdot P(B) + P(A|B^c) \cdot P(B^c)}$ In the case of the Bayesian network problem described above [P(B|A and D)] we have: $P(B|A \cap D) = \dfrac{P(A \cap D|B) \cdot P(B)}{P(A \cap D|B) \cdot P(B) + P(A \cap D|B^c) \cdot ...


1

What is the parameter $C$? Is it considered to be a random variable if it has probability density function $f_{C}$? Yes, $C$ is a random variable with PDF $f_C$ (when $C$ has a density). If so, what is its relationship to $(\Omega,\mathcal{A},\mu)$? Does it mean that $C:\Omega\to\mathbb{R}$? Yes, $C$ is defined on $\Omega$, if only to be able to ...


0

http://en.wikipedia.org/wiki/Bayesian_network: "A Bayesian network, Bayes network, belief network, Bayes(ian) model or probabilistic directed acyclic graphical model is a probabilistic graphical model (a type of statistical model) that represents a set of random variables and their conditional dependencies via a directed acyclic graph (DAG)." A DBN consist ...


0

No, $P(C|A,B)$ can't be determined solely from the marginal distributions $P(A)$ and $P(B)$. For example, say $C$ perfectly determines $A$ and $B$, via one of two mechanisms: If $C=1$, then $A=1$ and $B=1$ (and the marginal distribution of $C$ is the same as $A$ and $B$: $P(C=1) = 0.1$). In this case, $P(C=1|A=1, B=1)=1$. If $C=1$, then $A=0$ and $B=0$ ...


0

Non-Factorised: Say we have $3$ random variables for simplicity and that $r=2$: each variable having values $T$ or $F$. To know the full joint probability distribution, we need to know the actual probability value of each row in this table: $$ \begin{array}{ccc|c} x_1 & x_2 & x_3 & P(X_1=x_1,\; X_2=x_2,\; X_3=x_3) \\ \hline T & T & T ...


1

You have a prior distribution $$\pi_0(\theta) = \phi\left(\dfrac{\theta - \eta}{\sigma_0}\right)$$ and two likelihoods proportional to $\phi\left(\dfrac{s_1 - \theta}{\sigma_1}\right)$ and $\phi\left(\dfrac{s_2 - \theta}{\sigma_2}\right)$ so you want $$\pi(\theta|s_1,s_2)=\dfrac{\phi\left(\dfrac{\theta - \eta}{\sigma_0}\right)\phi\left(\dfrac{s_1 - ...


0

The key mistake is incorrectly writing the Binomial distribution pdf as $Pr(x_i|p_i,n_i)={n_i \choose x_i} x_i^{p_i} (n_i-x_i)^{1-p_i}$, instead of $Pr(x_i|p_i,n_i)={n_i \choose x_i} p_i^{x_i} (1-p_i)^{n_i-x_i}$. From here we can see our integrand is now of the form of a Beta distribution and thus integrates to $B(x_i + \alpha, n_i - x_i +\beta)$. ...


1

One thing I would change in the way you phrased the problem is to be sure you distinguish the event of getting black ball on the first draw from the event of getting a black ball on the second draw. These are two completely separate (though not independent) events. Clearer notation could help. You have to distinguish such things as "selected bin $A$ for the ...


2

Let $b_1$ be the event drawing a black ball the first draw; and $b_2$ the event of drawing a black ball the second draw. We want $P(b_2|b_1)$. That is, we want $\frac{P(b_2\cap b_1)}{P(b_1)}$. We have $P(b_1)=\frac12\cdot\frac12+\frac12\cdot\frac34=\frac58$. To figure out $P(b_1\cap b_2)$ we note that there are four ways this can happen: (i) Bin A, ...


0

Firstly, one should distinguish between the capital $X$ used to denote the random variable and the lower-case $x$ used as the argument to the density function or the cumulative distribution function, etc. This makes it possible to understand an expression like $\Pr(X\le x)=F(x)$ (with lower-case $x$ in two places and capital $X$ in one). Then $x\mapsto ...


0

If they would be conditionally independent, it would be possible to decompose $$P(A,B|C)=P(A|C)P(B|C)\textrm{.}$$ However, as you correctly write $$P(A,B|C) = \frac{P(A)P(B)P(C|A,B)}{P(C)}\textrm{.}$$ Thus, whether $A$ and $B$ are conditionally dependent depends on the decomposition of $P(C|A,B)=Q_1(A,C)Q_2(B,C)$ where $Q_1$ and $Q_2$ are some nonnegative ...



Top 50 recent answers are included