Bayes teoremi beklentileri karşılıyor mu?

İki rastgele değişken ve ,$A$$B$

$$E(A\mid B)=E(B\mid A)\frac{E(A)}{E(B)}?$$

$$E[A\mid B] \stackrel{?}= E[B\mid A]\frac{E[A]}{E[B]} \tag 1$$ Konjuge sonuç$(1)$,sıfır olmayan ortalamalara sahipbağımsızrastgele değişken$A$veBiçin önemsiz bir şekilde doğrudur$B$.

Eğer $E[B]=0$ , daha sonra sağ tarafı $(1)$ bir bölme içerir $0$ ve böylece $(1)$ bir anlamı yoktur. Olsun veya olmasın o Not $A$ ve $B$ bağımsız alakalı değildir.

Genel olarak , $(1)$ bağımlı rastgele değişkenler için geçerli değildir, ancak bağımlı $A$ ve $B$ tatmin eden ( 1 ) spesifik örnekleri $(1)$bulunabilir. E [ B ] ≠ 0 olduğu konusunda ısrar etmeye devam etmeliyiz $E[B]\neq 0$, aksi takdirde $(1)$ in sağ tarafı anlamsızdır. Ayı akılda $E[A\mid B]$ bir olan rastgele değişken bir olur işlevi rasgele değişkenin $B$ söylemek, $g(B)$ ise$E[B\mid A]$ a,rastgele değişkenbir birfonksiyonurastgele değişken bölgesinin$A$ , ki$h(A)$ . Yani,$(1)$ ,

$$g(B)\stackrel{?}= h(A)\frac{E[A]}{E[B]} \tag 2$$ gerçek bir açıklama olabilir, ve tabii ki cevabı olduğunu$g(B)$bir katı olamaz$h(A)$ genel olarak.

Bildiğim kadarıyla, $(1)$ tutabileceği sadece iki özel durum vardır .

• Yukarıda belirtildiği gibi, bağımsız rasgele değişkenler $A$ ve $B$ , $g(B)$ ve $h(A)$ , sırasıyla E [ A ] ve E [ B ] 'ye eşit olan dejenere rasgele değişkenlerdir (istatistiksel olarak okuma yazma bilmeyen kişiler tarafından sabitler olarak adlandırılır) ve eğer E [ B ] ≠ 0 , ( 1 ) ' de eşitliğe sahibiz .$E[A]$$E[B]$$E[B]\neq 0$$(1)$

• Bağımsızlık spektrum diğer ucunda, varsayalım ki $A=g(B)$ burada $g(\cdot)$ bir bir ters çevrilebilir fonksiyonu ve dolayısıyla $A=g(B)$ ve $B=g^{-1}(A)$ tamamen bağımlı rasgele değişkenlerdir. Bu durumda,

  $$E[A\mid B] = g(B), \quad E[B\mid A] = g^{-1}(A) = g^{-1}(g(B)) = B$$ ve böylece$(1)$ haline $$g(B)\stackrel{?}= B\frac{E[A]}{E[B]}$$ which holds exactly when $g(x) = \alpha x$ where $\alpha$ can be any nonzero real number. Thus, $(1)$ holds whenever $A$ is a scalar multiple of $B$, and of course $E[B]$ must be nonzero (cf. Michael Hardy's answer). The above development shows that $g(x)$ must be a linear function and that $(1)$ cannot hold for affine functions $g(x) = \alpha x + \beta$ with $\beta \neq 0$. However, note that Alecos Papadopolous in his answer and his comments thereafter claims that if $B$ is a normal random variable with nonzero mean, then for specific values of $\alpha$ and $\beta\neq 0$ that he provides, $A=\alpha B+\beta$ and $B$ satisfy $(1)$. In my opinion, his example is incorrect.

In a comment on this answer, Huber has suggested considering the symmetric conjectured equality

$$E[A\mid B]E[B] \stackrel{?}=E[B\mid A]E[A]\tag{3}$$ which of course always holds for independent random variables regardless of the values of $E[A]$ and $E[B]$ and for scalar multiples $A = \alpha B$ also. Of course, more trivially, $(3)$ holds for any zero-mean random variables $A$ and $B$ (independent or dependent, scalar multiple or not; it does not matter!): $E[A]=E[B]=0$ is sufficient for equality in $(3)$. Thus, $(3)$ might not be as interesting as $(1)$ as a topic for discussion.

The result is untrue in general, let us see that in a simple example. Let $X \mid P=p$ have a binomial distribution with parameters $n,p$ and $P$ have the beta distrubution with parameters $(\alpha, \beta)$, that is, a bayesian model with conjugate prior. Now just calculate the two sides of your formula, the left hand side is $\DeclareMathOperator{\E}{\mathbb{E}} \E X \mid P = nP$, while the right hand side is

$$\E( P\mid X) \frac{\E X}{\E P} = \frac{\alpha+X}{n+\alpha+\beta} \frac{\alpha/(\alpha+\beta)}{n\alpha/(\alpha+\beta)}$$ and those are certainly not equal.

The conditional expected value of a random variable $A$ given the event that $B=b$ is a number that depends on what number $b$ is. So call it $h(b).$ Then the conditional expected value $\operatorname{E}(A\mid B)$ is $h(B),$ a random variable whose value is completely determined by the value of the random variable $B$. Thus $\operatorname{E}(A\mid B)$ is a function of $B$ and $\operatorname{E}(B\mid A)$ is a function of $A$.

The quotient $\operatorname{E}(A)/\operatorname{E}(B)$ is just a number.

So one side of your proposed equality is determined by $A$ and the other by $B$, so they cannot generally be equal.

(Perhaps I should add that they can be equal in the trivial case when the values of $A$ and $B$ determine each other, as when for example, $A = \alpha B, \alpha \neq 0$ and $E[B]\neq 0$, when

$$E[A\mid B] = \alpha B = E[B\mid A]\cdot\alpha = E[B\mid A]\frac{\alpha E[B]}{E[B]} = E[B\mid A]\frac{E[A]}{E[B]}.$$ But functions equal to each other only at a few points are not equal.)

The expression certainly does not hold in general. For the fun of it, I show below that if $A$ and $B$ follow jointly a bivariate normal distribution, and have non-zero means, the result will hold if the two variables are linear functions of each other and have the same coefficient of variation (the ratio of standard deviation over mean) in absolute terms.

For jointly normals we have

$$\operatorname{E}(A \mid B) = \mu_A + \rho \frac{\sigma_A}{\sigma_B}(B - \mu_B)$$

and we want to impose

$$\mu_A + \rho \frac{\sigma_A}{\sigma_B}(B - \mu_B) = \left[\mu_B + \rho \frac{\sigma_B}{\sigma_A}(A - \mu_A)\right]\frac{\mu_A}{\mu_B}$$

$$\implies \mu_A + \rho \frac{\sigma_A}{\sigma_B}(B - \mu_B) = \mu_A + \rho \frac{\sigma_B}{\sigma_A}\frac{\mu_A}{\mu_B}(A - \mu_A)$$

Simplify $\mu_A$ and then $\rho$, and re-arrange to get

$$B = \mu_B +\frac{\sigma^2_B}{\sigma^2_A}\frac{\mu_A}{\mu_B}(A - \mu_A)$$

So this is the linear relationship that must hold between the two variables (so they are certainly dependent, with correlation coefficient equal to unity in absolute terms) in order to get the desired equality. What it implies?

First, it must also be satisfied that

$$E(B) \equiv \mu_B = \mu_B+\frac{\sigma^2_B}{\sigma^2_A}\frac{\mu_A}{\mu_B}(E(A) - \mu_A) \implies \mu_B = \mu_B$$

so no other restirction is imposed on the mean of $B$ ( or of $A$) except of them being non-zero. Also a relation for the variance must be satisfied,

$$\operatorname{Var}(B) \equiv \sigma^2_B = \left(\frac{\sigma^2_B}{\sigma^2_A}\frac{\mu_A}{\mu_B}\right)^2\operatorname{Var}(A)$$

$$\implies \left(\sigma^2_A\right)^2\sigma^2_B = \left(\sigma^2_B\right)^2\sigma^2_A\left(\frac{\mu_A}{\mu_B}\right)^2$$

$$\implies \left(\frac{\sigma_A}{\mu_A}\right)^2 = \left(\frac{\sigma_B}{\mu_B}\right)^2 \implies (\text{cv}_A)^2 = (\text{cv}_B)^2$$

$$\implies |\text{cv}_A| = |\text{cv}_B|$$

which was to be shown.

Note that equality of the coefficient of variation in absolute terms, allows the variables to have different variances, and also, one to have positive mean and the other negative.