Tag Info

New answers tagged

0

There is insufficient information to answer the question. What is missing is the relationship between the statement "the probability of catching a fish" and "the fisherman cast his rod three times and caught only one fish." Are we to assume, for example, that the probability of $0.4$ for pond $A$ is the per-casting probability, or the probability of ...


1

One interpretation of the problem is that every time she casts her rod say at Pond A, she has probability $0.4$ of catching a fish. We will assume, unreasonably, that the results of the casts are independent. Then given that she cast $3$ times in Pond A, the probability she caught exactly $1$ fish is $\binom{3}{1}(0.4)^1(0.6)^2$. Compute similar ...


0

The Bayes estimator $\lambda_B$ satisfies $\lambda_B = \arg \min_{\hat{\lambda}} \mathbb{E}(L(\hat{\lambda}, \lambda))$, that is, $\lambda_B$ is the value of $\hat{\lambda}$ that minimises the expected loss. So $$\lambda_{B} = \arg \min_{_\hat{\lambda}} \int_0^\infty |\hat{\lambda} - \lambda| p(\lambda | x_{1:5}) d \lambda.$$ Therefore $$\lambda_{B} = \arg ...


0

When you have causal relationships with variables influenced by their past and the past of other variables and there are cyclic characteristics, the problem has to be taken into a dynamic simulation structure using the most typical simulations methods depending on the problem you have to solve: - System Dynamics - Agent Based Models - Discrete Event Models ...


1

I think you're reading more into these equations than there is. This is just a disguised form of $$ p(y=1\mid x)=\frac{p(y=1\mid x)}{p(y=0\mid x)+p(y=1\mid x)}\;, $$ which is what you get if you plug the result you already obtained, $v(x)=\log\frac{p(y=1\mid x)}{p(y=0\mid x)}$, into the upper equation and multiply the fraction through by $p(y=1\mid x)$.


1

Probably you are given observations $X = \{X_1, \ldots, X_n\}$ with $X_i|\Theta = \theta \sim \text{ Bernoulli}(\theta)$, which gives the likelihood function of $\Theta$ as follows: $$\ell(x|\Theta = \theta) = \theta^{\sum_{i = 1}^n x_i}(1 - \theta)^{n - \sum_{i = 1}^n x_i} = \theta^t(1 - \theta)^{n - t},$$ where $t = \sum_{i = 1}^n x_i$. The prior of ...


2

I think what you call Bernoulli distribution should be called binomial distribution based on the context. Let \begin{align*} h(\theta)=\begin{cases}1&\text{if $\theta\in(0,1)$,}\\0&\text{otherwise}\end{cases} \end{align*} denote the prior probability density function of $\theta$ and $f(\theta\,|\,X)$ denote the posterior probability density ...


0

As you suggest, we should start by calculating $f(x_i, y_i, \pi)$. We can factorize it as $f(x_i, y_i, \pi) = f(y_i \mid x_i, \pi) f(x_i, \pi)$ by conditional probability, and we can make the following two observations: $f(y_i \mid x_i, \pi) = f(y_i \mid x_i) = g_{x_i}(y_i)$ as $Y$ is conditionally independent of $\pi$ given $X$. $f(x_i, \pi) = f(x_i \mid ...


0

Actually, if X and Y are disjoint. Then they are definitely $\color{red}{dependent}$. The reason is that if X happens, then Y will never happen. Maybe you should change your hypothesis about $X$, $Y$,and $Z$.


0

Is there some way to show that $P(a,b|e)=P(b|e)$ ? You are given that : $P(a\mid b, e) = 1$ Then as $P(a, b\mid e) = P(a\mid b, e) P(b\mid e)$ so...


2

I can add something to start an answer, but I'm no Java expert, sorry! $$\mathsf P(A) = \sum\limits_{b\in\{B,!B\}}\sum\limits_{e\in\{E,!E\}} P(A, b, e)$$ This is because of the partition rule of probability (more of an explanation is here ).   (Also known as the Total Probability Theorem, or Law of Total Probability.) Recall from conditional ...


0

I would go with conditional entropy: $H(A,B,C|readings)$, where $readings$ is the result of the measurements. Clearly, if you "read" all three instances, then no uncertainty is left, and indeed $$ H(A,B,C|A,B,C)=0. $$ Moreover, you have $H(A,B,C|A)=H(B,C|A)$. If you want to know which reading leaves you with the least uncertainty about the rest, then you ...


1

Your terminology is not exactly on the right track. Here is an example to show the likelihood function and how it is used in frequentist and Bayesian estimation. Bayesian and Frequentist Inference for Bernoulli Data Suppose we have a random sample $X_1, \dots, X_n$ from a certain population. In both frequentist and Bayesian approaches, the likelihood has ...


0

Even if the book/context uses a different P when there's a vertical line meaning "given", certainly the two Ps with vertical lines should be written the same. And all P's here can be be read "the probability of". As an aside, "Density function" is sort of ambiguous (cumulative density function? Probability density function only in the continuous case? ...


1

$P(C\vert\alpha,\beta)$ is not a density function, it's the probability of the event (or measurable set) $C$ conditioned by the random variables $\alpha$ and $\beta$.



Top 50 recent answers are included