New answers tagged

1

The statement $\mathbb P (Z|Y, X) = \mathbb P(Z|Y)$ means that $Z$ and $X$ are conditionally independent given $Y$ (a), which directly means $\mathbb P(X|Y,Z)=\mathbb P(X|Y)$ (which is (b)). So, the given statements are equivalent. But, the final statement $\mathbb P(Z|Y,X)=\mathbb P(X|Y,Z)$ is not always correct. You can see both using Bayes Rule, e,g. for ...


2

If you had known $P(A|B,C)$, then you could calculate $P(C|A,B)$ as follows: $$P(C|A,B)=\frac{P(A|B,C)P(B,C)}{P(A,B)}$$ You already know the joint probabilities $P(A,B)$ and $P(B,C)$ with what you have. Due to the usage of Bayes Rule, without a three-term (e.g. involving $A,B,C$ together) probability expression, it doesn't make sense to calculate another ...


0

One way to intuitively think of Bayes' theorem is that when any one of these is easy to calculate $$P(A∣B) ~~ \text{or } P(B∣A)$$ we can calculate the other one even though the other one seems to be bit hard at first Consider an example, Here $$P(A∣B)$$ is say I have a curtain and I told you there is an animal behind the curtain and given it is a four ...


1

I think there is a mistake in your proof when you define the event $A$ as $Y=y|X=x$, this definition does not make sense. You cannot include conditionality in an event (what would be a realization of such an event?), you can just talk about probability of an event conditionally to some other event. Conditioning on an event $X=x$ defines new probability ...


2

No, it's not required. Being $Y$ and $Z$ conditionally independent on $X$ directly implies $$P(Y|X)=P(Y|X,Z)$$ because $$P(Y|X,Z)=\frac{P(Y,Z|X)}{P(Z|X)}=\frac{\overbrace{P(Y|X)P(Z|X)}^{\text{conditional indep.}}}{P(Z|X)}=P(Y|X)$$


1

It's a bit messy, which might explain why it isn't seen more. Here is a sketch starting from the bivariate case, which generalizes. I'll use $X$ and $Y$ and $Z=X+Y.$ First let's find the conditional cdf for $Z$ given $X=x.$ $$F_{Z|X=x}=P \left[ X+Y \leq z \ | \ X=x\right]=P[Y \leq z-x]=F_Y(z-x)$$ Then the conditional pdf is found by differentiating: $$f_{...


1

The more reliable supplier is $A$ because its defection rate is smaller. And, we need the following probability: $$P(A|1D, 19D')=\frac{P(1D,19D'|A)P(A)}{P(1D,19D'|A)P(A)+P(1D,19D'|B)P(B)}$$ And, for example, one of the terms can be calculated as $$P(1D,19D'|A)={20 \choose 1}0.1^10.9^{19}$$ The rest can be calculated similarly and then substituted.


1

mutually exclusive means $P(A \cap B)= 0$. Hence \begin{align}P(A' \cap B')&=P((A \cup B)')\\&=1-P(A \cup B)\\&=1-P(A)-P(B)+P(A\cap B)\\ &= 1-P(A)-P(B) \end{align}


1

Intuitively, if you know that there are $y$ heads in $n$ tosses, no arrangement of 1s and 0s are superior to another, so each of the arrangements has the same probability, e.g. $$P(X^3=(1,1,0))=P(X^3=(1,0,1)=P(X^3=(0,1,1)))$$ which makes $$P(X^n|Y=y,\theta)=\frac{1}{n\choose y}$$ since there are $n \choose y$ equally likely situations. We can also find the ...


0

You could say that a statistic $T(x)$ is sufficient if you could model the sampling like first drawing the sufficient statistic from a distribution that depends on the parameters $$T \sim f(t\vert \theta)$$ and then draw the observations $X$ from a distribution that depends only on $T$ and is independent from $\theta$ $$X \vert T \sim g(x \vert T)$$ In ...


1

It all follows from the properties of multivariate normals. Since $X_i$ are independent and normally distributed, they're jointly normal, which means any linear combination of them is also jointly normal with them. So, $p_{\mathbf{X},Z_N}(\mathbf{x},z)$ is a multivariate normal, which in turn means $p_{X_i,Z_N}(x,z)$ is multivariate normal with $$\mu=\...


0

Hints: For $\theta$, \begin{align*} p(\theta|\tau,\textbf{y}) &\propto \overbrace{\tau^{\frac{n}{2} - 1} \exp [-\frac{\tau(n-1)}{2} s^2}^\text{does not depend on $\theta$} -\frac{\tau n}{2}(\bar{y} - \theta)^2 ]\\ &\propto \exp[-\frac{\tau n}{2}(\bar{y} - \theta)^2 ] \end{align*} and spot the Normal density in $\theta$ For $\tau$, \begin{align*} p(\...


2

Perhaps this could be addressed using a beta-binomial model? You could have a predictor that code for possible changes in the underlying (unknown) fraction of people with the disease in the population on the next day. This would give you a p-value, essentially telling you the probability of observing that data (on the next day) under the null hypothesis that ...


0

I think a simple way is using $Plug-in$ estimation. 1) estimate parameters of $\mu$ and $\Sigma$ of 3-variate normal based on data. $$(\hat{\mu} , \, \hat{\Sigma})$$ 2) Calculate the conditional distribution of $X_1|X_2,X_3$ and $X_1|X_2+X_3$ that depend on $\mu$ and $\Sigma$. 3) using $Plug-in$ estimation for $X_1|X_2,X_3$ and $X_1|X_2+X_3$ by ...


0

If $Y_1, ..., Y_n$ are iid $\text{Pois}(\lambda)$, then $T(Y) = \sum\limits_{i=1}^n Y_i \sim \text{Pois}(n\lambda)$ This is probably most easily seen by noting that the MGF for each $Y_i$ is $M_{y_i}(t) = e^{\lambda(e^t - 1)}$, so then the MGF for $T(Y)$ is $(M_{y_i}(t))^n = (e^{\lambda(e^t - 1)})^n = e^{(n\lambda)(e^t - 1)}$, which is the MGF for a $\text{...


1

There is an assumption that $t=T(y)$, otherwise the equation is not true. To see that the assumption is needed, suppose $t \ne T(y)$, then $P_\theta(Y=y,T(Y)=t)=0$ but $P(Y=y)$ can be positive. $$P(Y=y, T(Y)=t)=P(Y=y)$$ since we already know that $T(y)=t$. Also $$P(T(Y)=t)=\sum_{y_i:T(y_i)=t}P(Y=y_i)$$ since \begin{align}\{\omega: T(Y(\omega))=t\}&=\...


0

I was wondering if I could find the maxima of $P(w|y, \beta)$ by optimizing just $P(y|\beta, w) P(w)$. You can because you are doing a discriminate task and you don't need to care about the denominator which is a constant for every element you are going to compare. If you are doing a generative task you need to calculate the exact result of the $P(w|y, \...


0

To help a little bit more, case ii in your question is a special case that a sigma-algebra being conditioned on is generated by a partition. I am sure you will find that case intuitive as well.


3

One reason why causality cannot be expressed by conditional probability (without further assumptions) is that you can turn $P(A|B)$ into $P(B|A)$ with Bayes theorem and causality does not go both ways (similar to the saying „correlation does not imply causation“). An example in which neither $P(A|B)$ nor $P(B|A)$ give any causal information would be be ...


5

I agree with Alexis that this is a complicated question best explained by textbooks in causal inference like the ones mentioned. To briefly answer your question, the reason we cannot say that there is a causal effect of $B$ on $A$ is that there are other possible explanations that are consistent with the observed data. Getting the causal ordering of the ...


1

I believe it has to do with the Markovian assumption. It is assumed that $$P(X_n \mid X_{n-1}, \dots , X_0) = P(X_n \mid X_{n-1})$$ This is also true for larger lags $$P(X_n \mid X_{n-k}, X_{n-k-1}, \dots, X_0) = P(X_n \mid X_{n-k})$$


2

We are not given that $$P(A \cap B|Z)=P(A |Z)P(B|Z)$$ For example $A$ is the event that we get a head from tossing a fair coin. Let $Z$ be the event of getting a head from tossing another coin. Let $B$ be the event that the outcome of the two coin tosses being different. Then $A$ and $B$ are independent events. However, given $Z$, $A$ and $B$ are not the ...


1

Since this specification satisfies the Markov property, the simplest way to frame the problem is as a Markov chain. Let $X_t$ denote the value at "time" $t$, and note that it has the transition probability matrix: $$\mathbf{P} = \begin{bmatrix} 1-q & & q \\ 1-p & & p \\ \end{bmatrix}.$$ For simplicity, take the row labels and column labels ...


1

First write conditional probability formula: $$P(X_4=1|X_1=1)=\frac{P(X_1=1,X_4=1)}{P(X_1=1)}$$ Then, write the numerator in terms of the full joint: $$\begin{align}P(X_1=1,X_4=1)&=\sum_{x_2\in (0,1)}\sum_{x_3\in(0,1)}P(X_1=1,X_2=x_2,X_3=x_3,X_4=1)\end{align}$$ which can be calculated expanding the joint wrt graph provided. You can calculate $P(X_4)$ to ...


8

First of all, The computation is very theoretic and not a good representation or guideline for adapting your behavior (just in case, if that is what you are after). In the comments I had already mentioned several points of critique for this approach: The problem is that these computations will be based on highly subjective estimates about the underlying ...


0

It helps to use as much knowledge that you have about the problem. Geographical coordinates are usually 2D, longitude and latitude. If you are interested in the "spread" between predicted and actual landing site, you could take the total distance, e.g. the radius or calculate both differences in longitude and latitude. If you do this for all data points you ...


1

If you're asking for a rigorous definition of sufficiency using modern probability theory/measure theory, then one of the following definitions (in decreasing levels of generality) might be what you're looking for. Definition 1. (Sufficient sub-$\sigma$-algebra) Let $(\mathcal{X}, \mathcal{B})$ be a measurable space (the sample space), and let $\mathcal{P}$ ...


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