Attente de réciproque d'une variable

15

Je suis confus en appliquant l'attente au dénominateur.

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

peut-il être $1/E(X)\,$ ?

expected-value

— Shan
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Article connexe: stats.stackexchange.com/questions/305713/… .

— StubbornAtom

27

peut-il être 1 / E (X)?

Non, en général, ce n'est pas possible; L'inégalité de Jensen nous dit que si $X$ est une variable aléatoire et $\varphi$ est une fonction convexe, alors $\varphi(\text{E}[X]) \leq \text{E}\left[\varphi(X)\right]$ . Si $X$ est strictement positif, alors $1/X$ est convexe, donc $\text{E}[1/X]\geq 1/\text{E}[X]$ , et pour une fonction strictement convexe, l'égalité ne se produit que si $X$ a une variance nulle ... donc dans les cas où nous avons tendance à nous intéresser, les deux sont généralement inégaux.

En supposant que nous avons affaire à une variable positive, s'il est clair pour vous que $X$ et $1/X$ seront inversement liés ( $\text{Cov}(X,1/X)\leq 0$ ), cela impliquerait $E(X \cdot 1/X) - E(X) E(1/X) \leq 0$ ce qui implique $E(X) E(1/X) \geq 1$ , donc $E(1/X) \geq 1/E(X)$ .

Je suis confus en appliquant l'attente au dénominateur.

Utilisez la loi du statisticien inconscient

E [g (X)] = \int \infty - \infty g (x) f X (x) d X

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

(dans le cas continu)

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

Dans certains cas, l'attente peut être évaluée par inspection (par exemple avec des variables gamma aléatoires), ou en dérivant la distribution de l'inverse, ou par d'autres moyens.

— Glen_b -Reinstate Monica
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14

Comme Glen_b le dit, c'est probablement faux, car l'inverse est une fonction non linéaire. Si vous voulez une approximation de $E(1/X)$ peut - être que vous pouvez utiliser une expansion de Taylor autour $E(X)$ :

E (1 X) \approx E (1 E ( X ) - 1 E ( X ) 2 (X - E (X)) + 1 E ( X ) 3 (X - E (X)) 2) = = 1 E ( X ) + 1 E ( X ) 3 V a r (X)

$E \bigg( \frac{1}{X} \bigg) \approx E\bigg( \frac{1}{E(X)} - \frac{1}{E(X)^2}(X-E(X)) + \frac{1}{E(X)^3}(X - E(X))^2 \bigg) = \\ = \frac{1}{E(X)} + \frac{1}{E(X)^3}Var(X)$ so you just need mean and variance of X, and if the distribution of

X $X$ is symmetric this approximation can be very accurate.

EDIT: the maybe above is quite critical, see the comment from BioXX below.

— Matteo Fasiolo
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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

1/6f′′′(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

Xn−μ=Op(an) $X_n -\mu = O_p(a_n)$ where

an→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 (1 X) = \int \infty 0 x - 1 f (x) d x = \int \infty 0 (\int \infty 0 e - t x d t) f (x) d x = \int \infty 0 (\int \infty 0 e - t x f (x) d x) d t = \int \infty 0 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

MX(t)=11−t,t<1 $M_X(t)=\frac1{1-t}, t<1$ . Then

∫∞0MX(−t)dt=∫∞011+tdt=ln(1+t)∣∣∣∞0=∞ $\int_0^\infty M_X(-t)\; dt = \int_0^\infty \frac1{1+t} \; dt= \ln(1+t) \bigg\rvert_0^\infty = \infty$ , so definitely do not converge, and is very different from

1EX=11=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 \infty 0 e - λ x d λ = 1 x

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

\int \infty 0 E [e - λ X] d λ = E [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

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

tX $t X$ instead of

−tX $-t X$ 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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