Var (X) diketahui, bagaimana cara menghitung Var (1 / X)?

Jika saya hanya memiliki , bagaimana saya bisa menghitung ? $\mathrm{Var}(X)$ $\mathrm{Var}(\frac{1}{X})$

Saya tidak memiliki informasi tentang distribusi , jadi saya tidak dapat menggunakan transformasi, atau metode lain yang menggunakan distribusi probabilitas . $X$ $X$

distributions variance data-transformation

— ARAT
sumber

Saya pikir ini mungkin membantu Anda.

— Christoph_J

Itu tidak mungkin.

Pertimbangkan urutan $X_n$ dari variabel acak, di mana

P (X_{n} = n - 1) = P (X_{n} = n + 1) = 0.5

$P(X_n=n-1)=P(X_n=n+1)=0.5$

Kemudian:

V a r (X_{n}) = 1 for all n

$\newcommand{\Var}{\mathrm{Var}}\Var(X_n)=1 \quad \text{for all $n$}$

Tapi $\Var\left(\frac{1}{X_n}\right)$ mendekati nol ketika $n$ menuju tak terhingga:

V a r (\frac{1}{X_{n}}) = {(0.5 (\frac{1}{n + 1} - \frac{1}{n - 1}))}^{2}

$\Var\left(\frac{1}{X_n}\right)=\left(0.5\left(\frac{1}{n+1}-\frac{1}{n-1}\right)\right)^2$

Contoh ini menggunakan fakta bahwa $\Var(X)$ tidak berubah di bawah terjemahan $X$ , tetapi $\Var\left(\frac{1}{X}\right)$ tidak.

Tetapi bahkan jika kita asumsikan $\mathrm{E}(X)=0$ , kita tidak bisa menghitung $\Var\left(\frac{1}{X}\right)$ : Biarkan

P (X_{n} = - 1) = P (X_{n} = 1) = 0.5 (1 - \frac{1}{n})

$P(X_n=-1)=P(X_n=1)=0.5\left(1-\frac{1}{n}\right)$

dan

P (X_{n} = 0) = \frac{1}{n} for n > 0

$P(X_n=0)=\frac{1}{n} \quad \text{for $n>0$}$

Kemudian mendekati 1 ketika menuju infinity, tetapi $\Var(X_n)$ $n$ untuk semua. $\Var\left(\frac{1}{X_n}\right)=\infty$ $n$

— mogron
sumber

Anda bisa menggunakan deret Taylor untuk mendapatkan perkiraan momen orde rendah dari variabel acak yang ditransformasikan. Jika distribusi cukup 'ketat' di sekitar rata-rata (dalam arti tertentu), perkiraannya bisa sangat bagus.

Jadi misalnya

g (X) = g (μ) + (X - μ) g^{'} (μ) + \frac{(X - μ)^{2}}{2} g^{″} (μ) + \dots

$g(X) = g(\mu) + (X-\mu) g'(\mu) + \frac{(X-\mu)^2}{2} g''(\mu) + \ldots$

begitu

\begin{array}{rcl} Var [g (X)] & = & Var [g (μ) + (X - μ) g^{'} (μ) + \frac{(X - μ)^{2}}{2} g^{″} (μ) + \dots] \\ = & Var [(X - μ) g^{'} (μ) + \frac{(X - μ)^{2}}{2} g^{″} (μ) + \dots] \\ = & g^{'} (μ)^{2} Var [(X - μ)] + 2 g^{'} (μ) Cov [(X - μ), \frac{(X - μ)^{2}}{2} g^{″} (μ) + \dots] \\ + Var [\frac{(X - μ)^{2}}{2} g^{″} (μ) + \dots] \end{array}

$\begin{eqnarray} \text{Var}[g(X)] &=& \text{Var}[g(\mu) + (X-\mu) g'(\mu) + \frac{(X-\mu)^2}{2} g''(\mu) + \ldots]\\ &=& \text{Var}[(X-\mu) g'(\mu) + \frac{(X-\mu)^2}{2} g''(\mu) + \ldots]\\ &=& g'(\mu)^2 \text{Var}[(X-\mu)] + 2g'(\mu)\text{Cov}[(X-\mu),\frac{(X-\mu)^2}{2} g''(\mu) + \ldots] \\& &\quad+ \text{Var}[\frac{(X-\mu)^2}{2} g''(\mu) + \ldots]\\ \end{eqnarray}$

often only the first term is taken

Var [g (X)] \approx g^{'} (μ)^{2} Var (X)

$\text{Var}[g(X)] \approx g'(\mu)^2 \text{Var}(X)$

In this case (assuming I didn't make a mistake), with $g(X)=\frac{1}{X}$ , $\text{Var}[\frac{1}{X}] \approx \frac{1}{\mu^4} \text{Var}(X)$ .

Wikipedia: Taylor expansions for the moments of functions of random variables

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Some examples to illustrate this. I'll generate two (gamma-distributed) samples in R, one with a 'not-so-tight' distribution about the mean and one a bit tighter.

 a <- rgamma(1000,10,1)  # mean and variance 10; the mean is not many sds from 0
 var(a)
[1] 10.20819  # reasonably close to the population variance

The approximation suggests the variance of $1/a$ should be close to $(1/10)^4 \times 10 = 0.001$

 var(1/a)
[1] 0.00147171

Algebraic calculation has that the actual population variance is $1/648 \approx 0.00154$

Now for the tighter one:

 a <- rgamma(1000,100,10) # should have mean 10 and variance 1
 var(a)
[1] 1.069147

The approximation suggests the variance of $1/a$ should be close to $(1/10)^4 \times 1 = 0.0001$

 var(1/a)
[1] 0.0001122586

Algebraic calculation shows that the population variance of the reciprocal is $\frac{10^2}{99^2\times 98} \approx 0.000104$ .

— Glen_b -Reinstate Monica
sumber

Note that in this case, a quite weak hypothesis leads to the conclusion that no mean (whence variance) for

1 / X

$1/X$ will exist, i.e., that the approximation in the answer will be rather misleading. :-) An example hypothesis is that

X

$X$ has a density

f

$f$ that is continuous in an interval around zero and such that

f (0) \neq 0

$f(0) \neq 0$ . The result then follows because the density will be bounded away from zero on some interval

[- ϵ, ϵ]

$[-\epsilon,\epsilon]$ . The hypothesis just given is not the weakest possible, of course.

— cardinal

The reason the Taylor series argument then fails is because

\approx

$\approx$ hides the remainder (error) term, which in this case is

R (x, μ) = \frac{(x + μ) (x - μ)^{2}}{x μ},

$R(x,\mu) = \frac{(x+\mu)(x-\mu)^2}{x\mu} \>,$ and this behaves badly around

x = 0

$x = 0$ .

— cardinal

One must indeed be careful about the behavior of the density near 0. Note that in the above gamma examples, the distribution of the inverse is inverse gamma, for which having a finite mean requires

α > 1

$\alpha>1$ (

α

$\alpha$ being the shape parameter of the gamma we're inverting). The two examples had

α = 10

$\alpha = 10$ and

α = 100

$\alpha = 100$ . Even so (with "nice" distributions for inverting) neglect of higher terms can introduce a noticeable bias.

— Glen_b -Reinstate Monica

this seems in the right direction, of a reciprocal shifted normal distribution instead of a reciprocal standard normal distribution: en.wikipedia.org/wiki/…

— Felipe G. Nievinski