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

Jika saya hanya memiliki , bagaimana saya bisa menghitung ?V a r ( $\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$

Itu tidak mungkin.

Pertimbangkan urutan $X_n$ dari variabel acak, di mana

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

Kemudian:

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

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

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

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

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

dan

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

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

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

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

begitu

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

often only the first term is taken

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

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 $\frac{10^2}{99^2\times 98} \approx 0.000104$.