ベイズの定理は期待を満たしますか？

18

2つのランダム変数および、 $A$ $B$

E (A ∣ B) = E (B ∣ A) \frac{E (A)}{E (B)} ?

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

bayesian mathematical-statistics

— トムカ
ソース

3

うーん...私はこれらの2つの側面が同等だとは思わない

— ジョン

6

回答で指摘されているように、一方の確率変数が他方の条件変数であるため、この質問は確率的に無意味です。

— 西安

25

\begin{matrix} (1) & E [A ∣ B] \overset{?}{=} E [B ∣ A] \frac{E [A]}{E [B]} \end{matrix}

$E[A\mid B] \stackrel{?}= E[B\mid A]\frac{E[A]}{E[B]} \tag 1$ 推測された結果

(1)

$(1)$ は、非ゼロの平均を

独立したランダム変数

A

$A$ および

に対して自明です

B

$B$ 。

場合 $E[B]=0$ 、右側には $(1)$ による除算が含まれるため、は無意味です。とが独立しているかどうかは関係ないことに注意してください。 $0$ $(1)$ $A$ $B$

一般的に、従属確率変数が、従属の具体例については、保持していない及び満たす求めることができます。であると主張し続ける必要があることに注意してください。そうでなければ、右側は無意味です。ことを念頭に置いてベアである確率変数であることを起こる機能ランダム変数の、言う $(1)$ $A$ $B$ $(1)$ $E[B]\neq 0$ $(1)$ $E[A\mid B]$ $B$ 一方、である確率変数である関数ランダム変数の言う、。したがって、は、 $g(B)$ $E[B\mid A]$ $A$ $h(A)$ $(1)$

は真のステートメントである場合があり、明らかに答えは、一般にを倍数にすることはできないということです。

\begin{matrix} (2) & g (B) \overset{?}{=} h (A) \frac{E [A]}{E [B]} \end{matrix}

$g(B)\stackrel{?}= h(A)\frac{E[A]}{E[B]} \tag 2$

g (B)

$g(B)$

h (A)

$h(A)$

私の知る限り、が保持できる特別なケースは2つだけです。 $(1)$

上記のように、独立したランダム変数および場合、およびは、それぞれおよび等しい縮退ランダム変数（統計的に知識のない人々によって定数と呼ばれる）です。場合、は等式があります。 $A$ $B$ $g(B)$ $h(A)$ $E[A]$ $E[B]$ $E[B]\neq 0$ $(1)$
独立からのスペクトルのもう一方の端で、で、が可逆関数であり、したがっておよびが完全に従属する確率変数であると仮定します。この場合、 $A=g(B)$ $g(\cdot)$ $A=g(B)$ $B=g^{-1}(A)$ なので、は
$E [A ∣ B] = g (B), E [B ∣ A] = g^{- 1} (A) = g^{- 1} (g (B)) = B$ $E[A\mid B] = g(B), \quad E[B\mid A] = g^{-1}(A) = g^{-1}(g(B)) = B$ $(1)$ を正確保持している場合ここで任意のゼロでない実数であることができます。したがって、は、がスカラー倍数であり、もちろんがゼロ以外でなければならない場合に限ります（Michael Hardyの回答を参照）。上記の展開は、は線形関数でなければならず、はアフィンを保持できないことを示しています。 $g (B) \overset{?}{=} B \frac{E [A]}{E [B]}$ $g(B)\stackrel{?}= B\frac{E[A]}{E[B]}$ $g(x) = \alpha x$ $\alpha$ $(1)$ $A$ $B$ $E[B]$ $g(x)$ $(1)$ 関数と。しかし、ノートAlecosでPapadopolousその彼の答えと彼のコメントは、その後、場合という主張あり、通常の非ゼロの平均と確率変数、その後のための特定の 値と、彼が提供していることを、との関係を満たします。私の意見では、彼の例は間違っています。 $g(x) = \alpha x + \beta$ $\beta \neq 0$ $B$ $\alpha$ $\beta\neq 0$ $A=\alpha B+\beta$ $B$ $(1)$

この答えについてのコメントで、フーバーは対称推測平等考慮することを提案していどのコースに関係なく、常に値とは独立したランダム変数のために保持しているの及びとスカラー倍のためにも。もちろん、もっと簡単に、が成り立つ

\begin{matrix} (3) & E [A ∣ B] E [B] \overset{?}{=} E [B ∣ A] E [A] \end{matrix}

$E[A\mid B]E[B] \stackrel{?}=E[B\mid A]E[A]\tag{3}$

E [A]

$E[A]$

E [B]

$E[B]$

A = α B

$A = \alpha B$

(3)

$(3)$ 任意の ゼロ平均確率変数

および

（独立または従属、スカラー倍数またはそれ以外、重要ではありません！）：

は

等式に十分です。したがって、

は

ほど議論のトピックほど興味深いものではないかもしれません。

A

$A$

B

$B$

E [A] = E [B] = 0

$E[A]=E[B]=0$

(3)

$(3)$

(3)

$(3)$

(1)

$(1)$

— ディリップ・サルワテ
ソース

9

+1。寛大であるために、質問は

であるかどうかを尋ねると解釈できます。

E (A | B) E (B) = E (B | A) E (A)

$E(A|B)E(B)=E(B|A)E(A)$

— whuber

1

@whuber Thanks. My edit addresses the more general question as to whether it is possible to have

E [A ∣ B] E [B] = E [B ∣ A] E [A]

$E[A\mid B]E[B]=E[B\mid A]E[A]$ .

— Dilip Sarwate

11

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 ∣ X) \frac{E X}{E P} = \frac{α + X}{n + α + β} \frac{α / (α + β)}{n α / (α + β)}

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

— kjetil b halvorsen
ソース

2

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 ∣ B] = α B = E [B ∣ A] \cdot α = E [B ∣ A] \frac{α E [B]}{E [B]} = E [B ∣ A] \frac{E [A]}{E [B]} .

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

— Michael Hardy
ソース

You mean they are not necessarily equal? I mean they CAN be equal?

— BCLC

1

@BCLC : They are equal only in trivial cases. And two functions equal to each other at some points and not at others are not equal.

— Michael Hardy

2

"But only in that trivial case can they be equal" (emphasis added) is not quite correct. Consider independent

A

$A$ and

B

$B$ with

E [B] \neq 0

$E[B]\neq 0$ . Then,

E [A ∣ B] = E [A]

$E[A\mid B] = E[A]$ while

E [B ∣ A] = E [B]

$E[B\mid A] = E[B]$ and so

E [B ∣ A] \frac{E [A]}{E [B]} = E [B] \frac{E [A]}{E [B]} = E [A] = E [A ∣ B] .

$E[B\mid A] \frac{E[A]}{E[B]} = E[B]\frac{E[A]}{E[B]} = E[A] = E[A\mid B].$

— Dilip Sarwate

@DilipSarwate I was about to say that haha!

— BCLC

I edited your answer to add a few details for the case you pointed out. Please roll back if you don't like the changes.

— Dilip Sarwate

-1

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

E (A ∣ B) = μ_{A} + ρ \frac{σ_{A}}{σ_{B}} (B - μ_{B})

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

and we want to impose

μ_{A} + ρ \frac{σ_{A}}{σ_{B}} (B - μ_{B}) = [μ_{B} + ρ \frac{σ_{B}}{σ_{A}} (A - μ_{A})] \frac{μ_{A}}{μ_{B}}

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

⟹ μ_{A} + ρ \frac{σ_{A}}{σ_{B}} (B - μ_{B}) = μ_{A} + ρ \frac{σ_{B}}{σ_{A}} \frac{μ_{A}}{μ_{B}} (A - μ_{A})

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

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

B = μ_{B} + \frac{σ_{B}^{2}}{σ_{A}^{2}} \frac{μ_{A}}{μ_{B}} (A - μ_{A})

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

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 μ_{B} = μ_{B} + \frac{σ_{B}^{2}}{σ_{A}^{2}} \frac{μ_{A}}{μ_{B}} (E (A) - μ_{A}) ⟹ μ_{B} = μ_{B}

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

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,

Var (B) \equiv σ_{B}^{2} = {(\frac{σ_{B}^{2}}{σ_{A}^{2}} \frac{μ_{A}}{μ_{B}})}^{2} Var (A)

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

⟹ {(σ_{A}^{2})}^{2} σ_{B}^{2} = {(σ_{B}^{2})}^{2} σ_{A}^{2} {(\frac{μ_{A}}{μ_{B}})}^{2}

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

⟹ {(\frac{σ_{A}}{μ_{A}})}^{2} = {(\frac{σ_{B}}{μ_{B}})}^{2} ⟹ ({cv}_{A})^{2} = ({cv}_{B})^{2}

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

⟹ | {cv}_{A} | = | {cv}_{B} |

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

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.

— Alecos Papadopoulos
ソース

1

Isn't this a convoluted way to

A = α B

$A = \alpha B$ where

α

$\alpha$ is some scalar?

— Matthew Gunn

1

@MatthewGunn Your comment is right on target. Normality has nothing to do with the matter. For random variables

A

$A$ and

B

$B$ such that

A = α B

$A = \alpha B$ ,

E [A ∣ B] = α B = A

$E[A\mid B] = \alpha B = A$ and similarly,

E [B ∣ A] = B

$E[B\mid A] = B$ . Consequently, assuming that

E [B] \neq 0

$E[B]\neq 0$ ,

E [A ∣ B] = α B = E [B ∣ A] \cdot α = E [B ∣ A] \frac{α E [B]}{E [B]} = E [B ∣ A] \frac{E [A]}{E [B]} .

$E[A\mid B] = \alpha B = E[B\mid A]\cdot\alpha = E[B\mid A]\frac{\alpha E[B]}{E[B]} = E[B\mid A]\frac{E[A]}{E[B]}.$ No normality, no

| c v_{A} | = | c v_{B} |

$|cv_A|=|cv_B|$ etc, and actually just a rehash of a comment in Michael Hardy's answer.

— Dilip Sarwate

If you write \text{Var} instaed of \operatorname{Var} then you'll see

a Var X

$a\text{Var}X$ and

a Var (X)

$a\text{Var}(X)$ instead of

a Var X

$a\operatorname{Var}X$ and

a Var (X) .

$a\operatorname{Var}(X).$ That's why the latter is standard usage.

— Michael Hardy

@MatthewGun It seems to me that providing answers that contain specific examples is considered valuable content in this site. So yes, when a random variable is an affine function of another, and they are jointly normal with non-zero means, then one needs to have equal coefficients of variation, while, also there are no restrictions on the means of these rv's. On the other hand, when a random variable is just a linear function of another, the relation holds always. So no my answer is not a convoluted way to say

A = a B

$A=aB$ . (cc:@DilipSarwate)

— Alecos Papadopoulos

2

If

B

$B$ is a non-normal random variable with

E [B] = μ_{B} \neq 0

$E[B]=\mu_B\neq 0$ and

A = c B + d

$A=c B+d$ (and so

B = \frac{A - d}{c}

$B=\frac{A-d}{c}$ ), then

E [A ∣ B] = c B + d = A, E [B ∣ A] = \frac{A - d}{c} = B .

$E[A\mid B]=cB+d=A, E[B\mid A]=\frac{A-d}{c}=B.$ Now, if we want to have

E [A ∣ B] = c B + d

$E[A\mid B]=cB+d$ to equal

E [B ∣ A] \cdot \frac{μ_{A}}{μ_{B}} = B \cdot \frac{μ_{A}}{μ_{B}}

$E[B\mid A]\cdot\frac{\mu_A}{\mu_B} =B\cdot\frac{\mu_A}{\mu_B}$ , it must be that

c B + d = B \cdot \frac{μ_{A}}{μ_{B}} ⟹ d = 0, c = \frac{μ_{A}}{μ_{B}}

$cB+d=B\cdot\frac{\mu_A}{\mu_B}\implies d=0,c=\frac{\mu_A}{\mu_B}$ and so

A = c B = \frac{μ_{A}}{μ_{B}} B

$A=cB=\frac{\mu_A}{\mu_B}B$ . So, for nonnormal

B

$B$ , the OP's conjectured result holds if

A = c B

$A=cB$ but not if

A = c B + d, d \neq 0

$A=cB+d, d\neq 0$ .Of course, as you have proved, the result holds for normal random variables if

A = c B + d, d \neq 0

$A=cB+d, d\neq 0$ .

— Dilip Sarwate