की पीडीएफ

मान लीजिए कि से अज्ञात और $X_1, X_2,...,X_n$ $N(\mu,\sigma^2)$ $\mu \in \mathcal R$ $\sigma^2>0$

चलो एस मानक विचलन यहाँ है। $Z=\frac{X_1-\bar{X}}{S},$

यह दिखाया जा सकता है कि $Z$ में Lebesgue pdf है

f (z) = \frac{\sqrt{n} Γ (\frac{n - 1}{2})}{\sqrt{π} (n - 1) Γ (\frac{n - 2}{2})} {[1 - \frac{n z^{2}}{(n - 1)^{2}}]}^{n / 2 - 2} I_{(0, (n - 1) / \sqrt{n})} (| Z |)

$f(z)=\frac{\sqrt{n} \Gamma\left(\frac{n-1}{2}\right)}{\sqrt{\pi}(n-1)\Gamma\left(\frac{n-2}{2}\right)}\left[1-\frac{nz^2}{(n-1)^2}\right]^{n/2-2}I_{(0,(n-1)/\sqrt{n})}(|Z|)$

मेरा सवाल यह है कि इस pdf को कैसे प्राप्त करें?

सवाल से है यहाँ की UMVUE खोजने के लिए उदाहरण 3.3.4 में । मैं UMVUE को खोजने के लिए तर्क और प्रक्रियाओं को समझ सकता हूं, लेकिन पीडीएफ प्राप्त करने का तरीका नहीं जानता। $P(X_1 \le c)$

मुझे लगता है कि इस सवाल का भी इस से संबंधित एक

मदद के लिए बहुत बहुत धन्यवाद या किसी भी संबंधित संदर्भ को इंगित करने के लिए भी विनियोजित किया जाएगा।

self-study umvue

— दीप उत्तर
स्रोत

जवाबों:

इस परिणाम के बारे में कितना पेचीदा है यह एक सहसंबंध गुणांक के वितरण की तरह दिखता है। एक कारण है।

मान लीजिए शून्य सहसंबंध के साथ सामान्य है और दोनों चर के लिए सामान्य विचरण । एक iid नमूना ड्रा करें । यह सर्वविदित है, और आसानी से ज्यामितीय रूप से स्थापित (जैसा कि फिशर ने एक सदी पहले किया था) कि नमूना सहसंबंध गुणांक का वितरण $(X,Y)$ $\sigma^2$ $(x_1,y_1), \ldots, (x_n,y_n)$

r = \frac{\sum_{i = 1}^{n} (x_{i} - \bar{x}) (y_{i} - \bar{y})}{(n - 1) S_{x} S_{y}}

$r = \frac{\sum_{i=1}^n(x_i - \bar x)(y_i - \bar y)}{(n-1) S_x S_y}$

है

f (r) = \frac{1}{B (\frac{1}{2}, \frac{n}{2} - 1)} {(1 - r^{2})}^{n / 2 - 2}, - 1 \leq r \leq 1.

$f(r) = \frac{1}{B\left(\frac{1}{2}, \frac{n}{2}-1\right)}\left(1-r^2\right)^{n/2-2},\ -1 \le r \le 1.$

(यहाँ, हमेशा की तरह, और नमूना साधन हैं और और निष्पक्ष विचरण आकलनकर्ता के वर्ग जड़ें हैं।) है बीटा समारोह है, जिसके लिए $\bar x$ $\bar y$ $S_x$ $S_y$ $B$

\begin{matrix} (1) & \frac{1}{B (\frac{1}{2}, \frac{n}{2} - 1)} = \frac{Γ (\frac{n - 1}{2})}{Γ (\frac{1}{2}) Γ (\frac{n}{2} - 1)} = \frac{Γ (\frac{n - 1}{2})}{\sqrt{π} Γ (\frac{n}{2} - 1)} . \end{matrix}

$\frac{1}{B\left(\frac{1}{2}, \frac{n}{2}-1\right)} = \frac{\Gamma\left(\frac{n-1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{n}{2}-1\right)} = \frac{\Gamma\left(\frac{n-1}{2}\right)}{\sqrt{\pi}\Gamma\left(\frac{n}{2}-1\right)} . \tag{1}$

गणना करने के लिए , हम एक ही घुमाव के तहत नमूना के वितरण के चालान के साथ-साथ द्वारा बनाई गई रेखा के आसपास में घुमावों के तहत इसके आक्रमण का फायदा उठा सकते हैं , और चयन कर सकते हैं। किसी भी इकाई सदिश होने के लिए जिसके घटक शून्य के योग हैं। ऐसा एक सदिश आनुपातिक है । इसका मानक विचलन है $r$ $\mathbb{R}^n$ $(1,1,\ldots, 1)$ $y_i/S_y$ $v = (n-1, -1, \ldots, -1)$

S_{v} = \sqrt{\frac{1}{n - 1} ((n - 1)^{2} + (- 1)^{2} + \dots + (- 1)^{2})} = \sqrt{n} .

$S_v = \sqrt{\frac{1}{n-1}\left((n-1)^2 + (-1)^2 + \cdots + (-1)^2\right)} = \sqrt{n}.$

नतीजतन, के समान वितरण होना चाहिए $r$

\frac{\sum_{i = 1}^{n} (x_{i} - \bar{x}) (v_{i} - \bar{v})}{(n - 1) S_{x} S_{v}} = \frac{(n - 1) x_{1} - x_{2} - \dots - x_{n}}{(n - 1) S_{x} \sqrt{n}} = \frac{n (x_{1} - \bar{x})}{(n - 1) S_{x} \sqrt{n}} = \frac{\sqrt{n}}{n - 1} Z .

$\frac{\sum_{i=1}^n(x_i - \bar x)(v_i - \bar v)}{(n-1) S_x S_v} = \frac{(n-1)x_1 - x_2-\cdots-x_n}{(n-1) S_x \sqrt{n}} = \frac{n(x_1 - \bar x)}{(n-1) S_x \sqrt{n}} = \frac{\sqrt{n}}{n-1}Z.$

इसलिए हम सभी को के वितरण को खोजने के लिए को पुनर्विक्रय करना होगा : $r$ $Z$

f_{Z} (z) = | \frac{\sqrt{n}}{n - 1} | f (\frac{\sqrt{n}}{n - 1} z) = \frac{1}{B (\frac{1}{2}, \frac{n}{2} - 1)} \frac{\sqrt{n}}{n - 1} {(1 - \frac{n}{(n - 1)^{2}} z^{2})}^{n / 2 - 2}

$f_Z(z) = \big|\frac{\sqrt{n}}{n-1}\big| f\left(\frac{\sqrt{n}}{n-1}z\right) = \frac{1}{B\left(\frac{1}{2}, \frac{n}{2}-1\right)} \frac{\sqrt{n}}{n-1}\left(1- \frac{n}{(n-1)^2}z^2\right)^{n/2-2}$

$|z| \le \frac{n-1}{\sqrt{n}}$

$n=4$

$(x_i,y_i),i=1,\ldots,4$ while the second histogram plots the correlation coefficients of $(x_i,v_i),i=1,\ldots,4)$ for a randomly chosen vector $v_i$ that remains fixed for all iterations. They are both uniform. The QQ-plot on the right confirms these distributions are essentially identical.

Here's the R code that produced the plot.

n <- 4
n.sim <- 1e5
set.seed(17)
par(mfrow=c(1,3))
#
# Simulate spherical bivariate normal samples of size n each.
#
x <- matrix(rnorm(n.sim*n), n)
y <- matrix(rnorm(n.sim*n), n)
#
# Look at the distribution of the correlation of `x` and `y`.
#
sim <- sapply(1:n.sim, function(i) cor(x[,i], y[,i]))
hist(sim)
#
# Specify *any* fixed vector in place of `y`.
#
v <- c(n-1, rep(-1, n-1)) # The case in question
v <- rnorm(n)             # Can use anything you want
#
# Look at the distribution of the correlation of `x` with `v`.
#
sim2 <- sapply(1:n.sim, function(i) cor(x[,i], v))
hist(sim2)
#
# Compare the two distributions.
#
qqplot(sim, sim2, main="QQ Plot")

Reference

R. A. Fisher, Frequency-distribution of the values of the correlation coefficient in samples from an indefinitely large population. Biometrika, 10, 507. See Section 3. (Quoted in Kendall's Advanced Theory of Statistics, 5th Ed., section 16.24.)

— whuber
स्रोत

The link to the reference is broken.

— Sextus Empiricus

@Martijn Thank you for checking. I see what you mean--the link works, but it doesn't go to anything relevant! I have fixed it up.

— whuber

I'd like to suggest this way to get the pdf of Z by directly calculating the MVUE of $P(X\leq c)$ using Bayes' theorem although it's handful and complex.

Since $E[I_{(-\infty,c)}(X_1)]=P(X_1\leq c)$ and $Z_1=\bar X$ , $Z_2=S^2$ are joint complete sufficient statistic, MVUE of $P(X\leq c)$ would be like this:

ψ (z_{1}, z_{2}) = E [I_{(- \infty, c)} (X_{1}) | z_{1}, z_{2}] = \int_{- \infty}^{\infty} I_{(- \infty, c)} f_{X | Z_{1}, Z_{2}} (x_{1} | z_{1}, z_{2}) d x_{1}

$\psi(z_1,z_2)=E[I_{(-\infty,c)}(X_1)|z_1,z_2]=\int_{-\infty}^{\infty}I_{(-\infty,c)}f_{X|Z_1,Z_2}(x_1|z_1,z_2)dx_1$

Now using Bayes' theorem, we get

f_{X | Z_{1}, Z_{2}} (x_{1} | z_{1}, z_{2}) = \frac{f_{Z_{1}, Z_{2} | X_{1}} (z_{1}, z_{2} | x_{1}) f_{X_{1}} (x_{1})}{f_{Z_{1}, Z_{2}} (z_{1}, z_{2})}

$f_{X|Z_1,Z_2}(x_1|z_1,z_2)={{f_{Z_1,Z_2|X_1}(z_1,z_2|x_1)f_{X_1}(x_1)}\over{f_{Z_1,Z_2}(z_1,z_2)}}$

The denominator $f_{Z_1,Z_2}(z_1,z_2)=f_{Z_1}(z_1)f_{Z_2}(z_2)$ can be written in closed form because $Z_1 \sim N(\mu,\frac{\sigma^2}{n})$ , $Z_2 \sim \Gamma({n-1\over 2},{2 \sigma^2\over n-1})$ are independent of each other.

To get the closed form of numerator, we can adopt these statistics:

W_{1} = \frac{\sum_{i = 2}^{n} X_{i}}{n - 1}

$W_1 = {\sum_{i=2}^n X_i \over n-1}$

W_{2} = \frac{\sum_{i = 2}^{n} X_{i}^{2} - (n - 1) W_{1}^{2}}{(n - 1) - 1}

$W_2 = {\sum_{i=2}^n X_i^2 -(n-1) W_1^2 \over (n-1)-1}$

which is the mean and the sample variance of $X_2, X_3, ..., X_n$ and they are independent of each other and also independent of $X_1$ . We can express these in terms of $Z_1, Z_2$ .

$W_1={n Z_1 - X_1\over n-1}$ , $W_2={(n-1)Z_2+nZ_1^2-X_1^2-(n-1)W_1^2 \over n-2}$

We can use transformation while $X_1=x_1$ ,

f_{Z_{1}, Z_{2} | X_{1}} (z_{1}, z_{2} | x_{1}) = \frac{n}{n - 2} f_{W_{1}, W_{2}} (w_{1}, w_{2}) = \frac{n}{n - 2} f_{W_{1}} (w_{1}) f_{W_{2}} (w_{2})

$f_{Z_1,Z_2|X_1}(z_1,z_2|x_1)={n \over n-2}f_{W_1,W_2}(w_1,w_2)={n \over n-2}f_{W_1}(w_1)f_{W_2}(w_2)$

Since $W_1 \sim N(\mu,\frac{\sigma^2}{n-1})$ , $W_2 \sim \Gamma({n-2\over 2},{2 \sigma^2\over n-2})$ we can get the closed form of this. Note that this holds only for $w_2 \geq 0$ which restricts $x_1$ to $z_1-{n-1 \over \sqrt n}\sqrt{z_2} \leq x_1 \leq z_1+{n-1 \over \sqrt n}\sqrt{z_2}$ .

So put them all together, exponential terms would disappear and you'd get,

f_{X | Z_{1}, Z_{2}} (x_{1} | z_{1}, z_{2}) = \frac{Γ (\frac{n - 1}{2})}{\sqrt{π} Γ (\frac{n - 2}{2})} \frac{\sqrt{n}}{\sqrt{z_{2}} (n - 1)} (1 - {(\frac{\sqrt{n} (x_{1} - z_{1})}{\sqrt{z_{2}} (n - 1)})}^{2})

$f_{X|Z_1,Z_2}(x_1|z_1,z_2)={\Gamma({n-1 \over 2}) \over \sqrt{\pi} \Gamma({n-2 \over 2})} {\sqrt{n} \over \sqrt{z_2} (n-1)} (1-{({\sqrt{n} (x_1 -z_1) \over \sqrt{z_2} (n-1) })}^2)$ where

z_{1} - \frac{n - 1}{\sqrt{n}} \sqrt{z_{2}} \leq x_{1} \leq z_{1} + \frac{n - 1}{\sqrt{n}} \sqrt{z_{2}}

$z_1-{n-1 \over \sqrt n}\sqrt{z_2} \leq x_1 \leq z_1+{n-1 \over \sqrt n}\sqrt{z_2}$ and zero elsewhere.

From this,at this point, we can get the pdf of $Z={X_1- z_1 \over \sqrt{z_2}}$ using transformation.

By the way, the MVUE would be like this :

ψ (z_{1}, z_{2}) = \frac{Γ (\frac{n - 1}{2})}{\sqrt{π} Γ (\frac{n - 2}{2})} \int_{- \frac{π}{2}}^{θ_{c}} c o s^{n - 3} θ d θ

$\psi(z_1,z_2)={\Gamma({n-1 \over 2}) \over \sqrt{\pi} \Gamma({n-2 \over 2})} \int ^{\theta_c} _{-{\pi \over2}} cos^{n-3} \theta d\theta$ while

θ_{c} = s i n^{- 1} (\frac{\sqrt{n} (c - z_{1})}{(n - 1) \sqrt{z_{1}}})

$\theta_c = sin^{-1} ({\sqrt{n}(c-z_1)\over(n-1)\sqrt{z_1}})$ and would be 1 if

c \geq z_{1} + \frac{n - 1}{\sqrt{n} \sqrt{z_{2}}}

$c \geq z_1+{n-1 \over \sqrt{n} \sqrt{z_2} }$

I am not a native English speaker and there could be some awkward sentences. I am studying statistics by myself with text book introduction to mathmatical statistics by Hogg. So there could be some grammatical or mathmatical conceptual mistakes. It would be appreciated if someone correct them.

Thank you for reading.

— KDG
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