변수의 역수의 기대

15

분모에 기대를 적용하는 데 혼란 스럽습니다.

$E(1/X)=\,?$

수 있습니다 $1/E(X)\,$ ?

expected-value

— 샨
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관련 게시물 : stats.stackexchange.com/questions/305713/... .

— StubbornAtom

27

1 / E (X) 일 수 있습니까?

아니요, 일반적으로는 불가능합니다. 젠슨 부등식 우리에게 그 경우, $X$ 확률 변수이고, $\varphi$ 볼록 함수 인 다음 $\varphi(\text{E}[X]) \leq \text{E}\left[\varphi(X)\right]$ . 경우 $X$ 다음 엄격히 양의 $1/X$ , 그래서 볼록 $\text{E}[1/X]\geq 1/\text{E}[X]$ 및 엄격 볼록 함수, 평등에만 발생 $X$ 분산이 0입니다 ... 그래서 우리가 관심있는 경향이있는 경우, 두 개는 일반적으로 동일하지 않습니다.

우리가 양의 변수를 다루고 있다고 가정하면, $X$ 와 $1/X$ 가 반비례 관계 ( $\text{Cov}(X,1/X)\leq 0$ ) 라는 것이 분명하다면 이것은 $E(X \cdot 1/X) - E(X) E(1/X) \leq 0$ 의미 $E(X) E(1/X) \geq 1$ 따라서 입니다. $E(1/X) \geq 1/E(X)$

분모에 기대를 적용하는 데 혼란 스럽습니다.

무의식 통계학 자의 법칙을 사용하라

E [g (X)] = \int_{- \infty}^{\infty} g (x) f_{X} (x) d x

$\text{E}[g(X)] = \int_{-\infty}^\infty g(x) f_X(x) dx$

(연속적인 경우)

따라서 , $g(X) = \frac{1}{X}$ $\text{E}[\frac{1}{X}]=\int_{-\infty}^\infty \frac{f(x)}{x} dx$

어떤 경우에는 검사를 통해 (예 : 감마 랜덤 변수를 사용하여) 또는 역 분포를 유도하거나, 다른 방법으로 기대치를 평가할 수 있습니다.

— Glen_b-복귀 모니카
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14

Glen_b가 말했듯이, 역수는 비선형 함수이기 때문에 아마도 잘못된 것입니다. 당신이에 근사 할 경우 $E(1/X)$ 어쩌면 당신은 주위 테일러 전개 사용할 수 있습니다 $E(X)$ :

E (\frac{1}{X}) \approx E (\frac{1}{E (X)} - \frac{1}{E (X)^{2}} (X - E (X)) + \frac{1}{E (X)^{3}} (X - E (X))^{2}) = = \frac{1}{E (X)} + \frac{1}{E (X)^{3}} V a r (X)

$E \bigg( \frac{1}{X} \bigg) \approx E\bigg( \frac{1}{E(X)} - \frac{1}{E(X)^2}(X-E(X)) + \frac{1}{E(X)^3}(X - E(X))^2 \bigg) = \\ = \frac{1}{E(X)} + \frac{1}{E(X)^3}Var(X)$ 그냥 의미와 X의 분산하고, 필요 그래서 분포의 경우

X

$X$ 대칭이 근사는 매우 정확이 될 수 있습니다.

편집 : 아마도 위의 내용은 매우 중요합니다. 아래 BioXX의 의견을 참조하십시오.

— 마테오 파실로
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oh yes yes...I am very sorry that I could not apprehend that fact...I have one more q...Is this applicable to any kind of function???actually I am stuck with

| x |

$|x|$ ...How can the expectation of

| x |

$|x|$ can be deduced in terms of

E (x)

$E(x)$ and

V (x)

$V(x)$

— Sandipan Karmakar

2

I don't think you can use it for

| X |

$|X|$ as that function is not differentiable. I would rather divide the problem into the cases and say

E (| X |) = E (X | X > 0) p (X > 0) + E (- X | X < 0) p (X < 0)

$E(|X|) = E(X|X > 0)p(X>0) + E(-X|X < 0)p(X<0)$ , I guess.

— Matteo Fasiolo

1

@MatteoFasiolo Can you please explain why the symmetry of the distribution of

X

$X$ (or lack thereof) has an effect on the accuracy of the Taylor approximation? Do you have a source that you could point me to that explains why this is?

— Aaron Hendrickson

1

@AaronHendrickson my reasoning is simply that the next term in the expansion is proportional to

E {(X - E (X))^{3}}

$E\{(X-E(X))^3\}$ which is related to the skewness of the distribution of

X

$X$ . Skewness is an asymmetry measure. However, zero skewness does not guarantee symmetry and I am not sure whether symmetry guarantees zero skewness. Hence, this is all heuristic and there might be plenty of counterexamples.

— Matteo Fasiolo

4

I don't understand how this solution gets so many upvotes. For a single random variable

X

$X$ there is no justificiation about the quality of this approximation. The third derivative

f (x) = 1 / x

$f(x)=1/x$ is not bounded. Moreover the remainder of the approx. is

1 / 6 f^{‴} (ξ) (X - μ)^{3}

$1/6f'''(\xi)(X-\mu)^3$ where

ξ

$\xi$ is itself a random variable between

X

$X$ and

μ

$\mu$ . The remainder won't vanish in general and may be very huge. Taylor approx. may only be useful if one has sequence of random variables

X_{n} - μ = O_{p} (a_{n})

$X_n -\mu = O_p(a_n)$ where

a_{n} \to 0

$a_n \to 0$ . Even then uniform integrability is needed additionally if interested in the expectation.

— BloXX

8

Others have already explained that the answer to the question is NO, except trivial cases. Below we give an approach to finding $\DeclareMathOperator{\E}{\mathbb{E}} \E \frac1{X}$ when $X>0$ with probability one, and the moment generating function $M_X(t) = \E e^{tX}$ do exist. An application of this method (and a generalization) is given in Expected value of $1/x$ when $x$ follows a Beta distribution, we will here also give a simpler example.

First, note that $\int_0^\infty e^{-t x}\; dt = \frac1{x}$ (simple calculus exercise). Then, write

E (\frac{1}{X}) = \int_{0}^{\infty} x^{- 1} f (x) d x = \int_{0}^{\infty} (\int_{0}^{\infty} e^{- t x} d t) f (x) d x = \int_{0}^{\infty} (\int_{0}^{\infty} e^{- t x} f (x) d x) d t = \int_{0}^{\infty} M_{X} (- t) d t

$\E \left(\frac1{X}\right) = \int_0^\infty x^{-1} f(x)\; dx = \int_0^\infty \left( \int_0^\infty e^{-tx}\; dt \right) f(x)\; dx =\\ \int_0^\infty \left( \int_0^\infty e^{-tx} f(x) \; dx \right) \; dt = \int_0^\infty M_X(-t) \; dt$ A simple application: Let

X

$X$ have the exponential distribution with rate 1, that is, with density

e^{- x}, x > 0

$e^{-x}, x>0$ and moment generating function

M_{X} (t) = \frac{1}{1 - t}, t < 1

$M_X(t)=\frac1{1-t}, t<1$ . Then

\int_{0}^{\infty} M_{X} (- t) d t = \int_{0}^{\infty} \frac{1}{1 + t} d t = \ln (1 + t) |_{0}^{\infty} = \infty

$\int_0^\infty M_X(-t)\; dt = \int_0^\infty \frac1{1+t} \; dt= \ln(1+t) \bigg\rvert_0^\infty = \infty$ , so definitely do not converge, and is very different from

\frac{1}{E X} = \frac{1}{1} = 1

$\frac1{\E X}=\frac11=1$ .

— kjetil b halvorsen
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7

An alternative approach to calculating $E(1/X)$ knowing X is a positive random variable is through its moment generating function $E[e^{-\lambda X}]$ . Since by elementary calculas

\int_{0}^{\infty} e^{- λ x} d λ = \frac{1}{x}

$\int_0^\infty e^{-\lambda x} d\lambda =\frac{1}{x}$ we have, by Fubini's theorem

\int_{0}^{\infty} E [e^{- λ X}] d λ = E [\frac{1}{X}] .

$\int_0^\infty E[e^{-\lambda X}] d\lambda =E[\frac{1}{X}].$

— user172761
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2

The idea here is right, but the details wrong. Pleasecheck

— kjetil b halvorsen

1

@Kjetil I don't see what the problem is: apart from the inconsequential differences of using

t X

$t X$ instead of

- t X

$-t X$ in the definition of the MGF and naming the variable

t

$t$ instead of

λ

$\lambda$ , the answer you just posted is identical to this one.

— whuber

1

You are right, the problems was less than I thought. Still this answer would be better withm some more details. I will upvote this tomorrow ( when I have new votes)

— kjetil b halvorsen

1

To first give an intuition, what about using the discrete case in finite sample to illustrate that $\text{E}(1/X)\neq 1/\text{E}(X)$ (putting aside cases such as $\text{E}(X)=0$ )?

In finite sample, using the term average for expectation is not that abusive, thus if one has on the one hand

$\text{E}(X) = \frac{1}{N}\sum_{i=1}^N X_i$

and one has on the other hand

$\text{E}(1/X) = \frac{1}{N}\sum_{i=1}^N 1/X_i$

it becomes obvious that, with $N>1$ ,

$\text{E}(1/X) = \frac{1}{N}\sum_{i=1}^N 1/X_i \neq \frac{N}{\sum_{i=1}^N X_i} = 1/\text{E}(X)$

Which leads to say that, basically, $\text{E}(1/X)\neq 1/\text{E}(X)$ since the inverse of the (discrete) sum is not the (discrete) sum of inverses.

Analogously in the asymptotic $0$ -centered continuous case, one has

$\text{E}(1/X)=\int_{-\infty}^\infty \frac{f(x)}{x} dx \neq 1/\int_{-\infty}^\infty xf(x) dx = 1/\text{E}(X)$ .

— keepAlive
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