এটি একটি খুব বেসিক প্রশ্ন। কেন আমরা চি চি স্কোয়ার বিতরণ ব্যবহার করব? এই বিতরণের অর্থ কী? কেন এই বিতরণ বৈকল্পিকের জন্য একটি আস্থা অন্তর তৈরি করতে ব্যবহৃত হয়?

আমি গুগল করার জন্য প্রতিটি জায়গাতেই এটি একটি সত্য উপস্থাপন করে, চি কখন ব্যবহার করতে হবে তা ব্যাখ্যা করে, তবে কেন চি ব্যবহার করতে হবে এবং কেন এটি দেখতে যেভাবে দেখায় তা ব্যাখ্যা করে না।

যে কেউ আমাকে সঠিক দিকের দিকে নির্দেশ করতে পারে এবং এটির জন্য অনেক ধন্যবাদ - যখন আমি বৈকল্পিকতার জন্য একটি আত্মবিশ্বাসের ব্যবধান তৈরি করছি তখন কেন আমি চি ব্যবহার করছি understanding

variance chi-squared

— nafrtiti
সূত্র

আপনি এটি ব্যবহার করেন কারণ - যখন ডেটা স্বাভাবিক থাকে -

Q = (n - 1) \frac{s^{2}}{σ^{2}} \sim χ_{n - 1}^{2}

$Q = (n-1)\frac{s^2}{\sigma^2}\sim \chi^2_{n-1}$ । (এটি

Q

$Q$ একটি মৌলিক পরিমাণে পরিণত করে)

— Glen_b -Rininateate Monica

এছাড়াও stats.stackexchange.com/questions/15711/… এবং এর লিঙ্ক দেখুন।

— নিক কক্স

যারা বা আরো গবেষণার অ্যাপ্লিকেশন করার জন্য আগ্রহী

, আপনি একটি মধ্যে পার্থক্য দিতে মনোযোগ করতে চাইবেন

( "চি-স্কোয়ারড") বিতরণ এবং একটি

( "চি") বন্টন (এটা আশ্চর্যজনকভাবে একটি

বর্গমূল )

χ^{2}

$\chi^2$

χ^{2}

$\chi^2$

χ

$\chi$

χ^{2}

$\chi^2$

— হোবার

Quick answer

The reason is because, assuming the data are i.i.d. and $X_i\sim N(\mu,\sigma^2)$ , and defining

\begin{array}{rcl} \bar{X} & = & \sum^{N} \frac{X_{i}}{N} \\ S^{2} & = & \sum^{N} \frac{(\bar{X} - X_{i})^{2}}{N - 1} \end{array}

$\begin{eqnarray*} \bar{X}&=&\sum^N \frac{X_i}{N}\\ S^2 &=& \sum^{N} \frac{(\bar{X}-X_i)^2}{N-1} \end{eqnarray*}$ when forming confidence intervals, the sampling distribution associated with the sample variance (

S^{2}

$S^2$ , remember, a random variable!) is a chi-square distribution (

S^{2} (N - 1) / σ^{2} \sim χ_{n - 1}^{2}

$S^2(N-1)/\sigma^2 \sim \chi^2_{n-1}$ ), just as the sampling distribution associated with the sample mean is a standard normal distribution (

(\bar{X} - μ) \sqrt{n} / σ \sim Z (0, 1)

$(\bar{X}-\mu)\sqrt{n}/\sigma \sim Z(0,1)$ ) when you know the variance, and with a t-student when you don't (

(\bar{X} - μ) \sqrt{n} / S \sim T_{n - 1}

$(\bar{X}-\mu)\sqrt{n}/S \sim T_{n-1}$ ).

Long answer

First of all, we'll prove that $S^2(N-1)/\sigma^2$ follows a chi-square distribution with $N-1$ degrees of freedom. After that, we'll see how this proof is useful when deriving the confidence intervals for the variance, and how the chi-square distribution appears (and why it is so useful!). Let's begin.

The proof

$\nu$

m_{χ_{ν}^{2}} (t) = (1 - 2 t)^{- ν / 2} .

$\begin{equation*} m_{\chi^2_\nu}(t)=(1-2t)^{-\nu/2}. \end{equation*}$ If we can show that the distribution of

S^{2} (N - 1) / σ^{2}

$S^2(N-1)/\sigma^2$ has a moment generating function like this one, but with

ν = N - 1

$\nu=N-1$ , then we have shown that

S^{2} (N - 1) / σ^{2}

$S^2(N-1)/\sigma^2$ follows a chi-square distribution with

N - 1

$N-1$ degrees of freedom. In order to show this, note two facts:

If we define,
$Y = \sum \frac{(X_{i} - \bar{X})^{2}}{σ^{2}} = \sum Z_{i}^{2},$ $\begin{equation*} Y = \sum \frac{(X_i-\bar{X})^2}{\sigma^2} = \sum Z_i^2, \end{equation*}$ where $Z_i\sim N(0,1)$ , i.e., standard normal random variables, the moment generating function of $Y$ is given by $\begin{array}{rcl} m_{Y} (t) & = & E [e^{t Y}] \\ = & E [e^{t Z_{1}^{2}}] \times E [e^{t Z_{2}^{2}}] \times . . . E [e^{t Z_{N}^{2}}] \\ = & m_{Z_{i}^{2}} (t) \times m_{Z_{2}^{2}} (t) \times . . . m_{Z_{N}^{2}} (t) . \end{array}$ $\begin{eqnarray*} m_Y(t) &=& \mathbb{E}[e^{tY}]\\ &=&\mathbb{E}[e^{tZ_1^2}]\times \mathbb{E}[e^{tZ_2^2}]\times ...\mathbb{E}[e^{tZ_N^2}]\\ &=&m_{Z_i^2}(t)\times m_{Z_2^2}(t)\times ...m_{Z_N^2}(t). \end{eqnarray*}$ The MGF of $Z^2$ is given by $\begin{array}{rcl} m_{Z^{2}} (t) & = & \int_{- \infty}^{\infty} f (z) \exp (t z^{2}) d z \\ = & (1 - 2 t)^{- 1 / 2}, \end{array}$ $\begin{eqnarray*} m_{Z^2}(t) &=& \int_{-\infty}^{\infty} f(z)\exp(tz^2)dz\\ &=&(1-2t)^{-1/2}, \end{eqnarray*}$ where I have used the PDF of the standard normal, $f(z)=e^{-z^2/2}/\sqrt{2\pi}$ and, hence, $m_{Y} (t) = (1 - 2 t)^{- N / 2},$ $\begin{equation*} m_Y(t)=(1-2t)^{-N/2}, \end{equation*}$ which implies that $Y$ follows a chi-square distribution with $N$ degrees of freedom.
If $Y_1$ and $Y_2$ are independent and each distribute as a chi-square distribution but with $\nu_1$ and $\nu_2$ degrees of freedom, then $W=Y_1+Y_2$ distributes with a chi-square distribution with $\nu_1+\nu_2$ degrees of freedom (this follows from taking the MGF of $W$ ; do this!).

With the above facts, note that if you multiply the sample variance by $N-1$ , you obtain (after some algebra),

(N - 1) S^{2} = - n (\bar{X} - μ) + \sum (X_{i} - μ)^{2},

$\begin{equation*} (N-1)S^2 = -n(\bar{X}-\mu)+\sum(X_i-\mu)^2, \end{equation*}$ and, hence, dividing by

σ^{2}

$\sigma^2$ ,

\frac{(N - 1) S^{2}}{σ^{2}} + \frac{(\bar{X} - μ)^{2}}{σ^{2} / N} = \sum \frac{(X_{i} - μ)^{2}}{σ^{2}} .

$\begin{equation*} \frac{(N-1)S^2}{\sigma^2}+\frac{(\bar{X}-\mu)^2}{\sigma^2/N}=\sum \frac{(X_i-\mu)^2}{\sigma^2}. \end{equation*}$ Note that the second term in the left-side of this sum distributes as a chi-square distribution with 1 degree of freedom, and the right-hand side sum distributes as a chi-square with

N

$N$ degrees of freedom. Therefore, $S^2(N-1)/\sigma^2$ distributes as a chi-square with $N-1$ degrees of freedom.

Calculating the Confidence Interval for the variance.

When looking for a confidence interval for the variance, you want to know the limits $L_1$ and $L_2$ in

P (L_{1} \leq σ^{2} \leq L_{2}) = 1 - α .

$\begin{equation*} \mathbb{P}\left(L_1\leq \sigma^2 \leq L_2\right) = 1-\alpha. \end{equation*}$ Let's play with the inequality inside the parenthesis. First, divide by

S^{2} (N - 1)

$S^2(N-1)$ ,

\frac{L_{1}}{S^{2} (N - 1)} \leq \frac{σ^{2}}{S^{2} (N - 1)} \leq \frac{L_{2}}{S^{2} (N - 1)} .

$\begin{equation*} \frac{L_1}{S^2(N-1)}\leq \frac{\sigma^2}{S^2(N-1)} \leq \frac{L_2}{S^2(N-1)}. \end{equation*}$ And then remember two things: (1) the statistic

S^{2} (N - 1) / σ^{2}

$S^2(N-1)/\sigma^2$ has a chi-squared distribution with

N - 1

$N-1$ degrees of freedom and (2) the variances is always greather than zero, which implies that you can invert the inequalities, because

\begin{array}{rcl} \frac{L_{1}}{S^{2} (N - 1)} \leq \frac{σ^{2}}{S^{2} (N - 1)} & \Rightarrow & \frac{S^{2} (N - 1)}{σ^{2}} \leq \frac{S^{2} (N - 1)}{L_{1}}, \\ \frac{σ^{2}}{S^{2} (N - 1)} \leq \frac{L_{2}}{S^{2} (N - 1)} & \Rightarrow & \frac{S^{2} (N - 1)}{L_{2}} \leq \frac{S^{2} (N - 1)}{σ^{2}}, \end{array}

$\begin{eqnarray*} \frac{L_1}{S^2(N-1)}\leq \frac{\sigma^2}{S^2(N-1)} &\Rightarrow& \frac{S^2(N-1)}{\sigma^2}\leq \frac{S^2(N-1)}{L_1},\\ \frac{\sigma^2}{S^2(N-1)} \leq \frac{L_2}{S^2(N-1)} &\Rightarrow& \frac{S^2(N-1)}{L_2} \leq \frac{S^2(N-1)}{\sigma^2},\\ \end{eqnarray*}$ hence, the probability we are looking for is:

P (\frac{S^{2} (N - 1)}{L_{2}} \leq \frac{S^{2} (N - 1)}{σ^{2}} \leq \frac{S^{2} (N - 1)}{L_{1}}) = 1 - α .

$\begin{equation*} \mathbb{P}\left(\frac{S^2(N-1)}{L_2} \leq \frac{S^2(N-1)}{\sigma^2}\leq \frac{S^2(N-1)}{L_1}\right) = 1-\alpha. \end{equation*}$ Note that

S^{2} (N - 1) / σ^{2} \sim χ^{2} (N - 1)

$S^2(N-1)/\sigma^2 \sim \chi^2(N-1)$ . We want then,

\begin{array}{rcl} \int_{\frac{S^{2} (N - 1)}{L_{2}}}^{N - 1} p_{χ^{2}} (x) d x & = & (1 - α) / 2, \\ \int_{N - 1}^{\frac{S^{2} (N - 1)}{L_{1}}} p_{χ^{2}} (x) d x & = & (1 - α) / 2 \end{array}

$\begin{eqnarray*} \int_{\frac{S^2(N-1)}{L_2}}^{N-1}p_{\chi^2}(x)dx &=& (1-\alpha)/2\ \ \ ,\\ \int_{N-1}^{\frac{S^2(N-1)}{L_1}}p_{\chi^2}(x)dx &=& (1-\alpha)/2\ \ \, \end{eqnarray*}$ (we integrate up to

N - 1

$N-1$ because the expected value of a chi-squared random variable with

N - 1

$N-1$ degrees of freedom is

N - 1

$N-1$ ) or, equivalently,

\begin{array}{rcl} \int_{0}^{\frac{S^{2} (N - 1)}{L_{2}}} p_{χ^{2}} (x) d x = α / 2, \\ \int_{\frac{S^{2} (N - 1)}{L_{1}}}^{\infty} p_{χ^{2}} (x) d x = α / 2. \end{array}

$\begin{eqnarray*} \int_{0}^{\frac{S^2(N-1)}{L_2}}p_{\chi^2}(x)dx=\alpha/2,\\ \int_{\frac{S^2(N-1)}{L_1}}^{\infty}p_{\chi^2}(x)dx=\alpha/2. \end{eqnarray*}$ Calling

χ_{α / 2}^{2} = \frac{S^{2} (N - 1)}{L_{2}}

$\chi^2_{\alpha/2}=\frac{S^2(N-1)}{L_2}$ and

χ_{1 - α / 2}^{2} = \frac{S^{2} (N - 1)}{L_{1}}

$\chi^2_{1-\alpha/2}= \frac{S^2(N-1)}{L_1}$ , where the values

χ_{α / 2}^{2}

$\chi^2_{\alpha/2}$ and

χ_{1 - α / 2}^{2}

$\chi^2_{1-\alpha/2}$ can be found in chi-square tables (in computers mainly!) and solving for

L_{1}

$L_1$ and

L_{2}

$L_2$ ,

\begin{array}{rcl} L_{1} & = & \frac{S^{2} (N - 1)}{χ_{1 - α / 2}^{2}}, \\ L_{2} & = & \frac{S^{2} (N - 1)}{χ_{α / 2}^{2}} . \end{array}

$\begin{eqnarray*} L_1 &=& \frac{S^2(N-1)}{\chi^2_{1-\alpha/2}},\\ L_2 &=& \frac{S^2(N-1)}{\chi^2_{\alpha/2}}. \end{eqnarray*}$ Hence, your confidence interval for the variance is

C . I . = (\frac{S^{2} (N - 1)}{χ_{1 - α / 2}^{2}}, \frac{S^{2} (N - 1)}{χ_{α / 2}^{2}}) .

$\begin{equation*} C.I.=\left(\frac{S^2(N-1)}{\chi^2_{1-\alpha/2}}, \frac{S^2(N-1)}{\chi^2_{\alpha/2}}\right). \end{equation*}$

— Néstor
সূত্র

Simply because

S^{2}

$S^2$ does not follow a centered chi-square distribution, while

S^{2} (N - 1) / σ^{2}

$S^2(N-1)/\sigma^2$ does and, therefore, its easier to work with. Are you asking for a derivation for that? (i.e., you want someone to show you that

S^{2} (N - 1) / σ^{2}

$S^2(N-1)/\sigma^2$ follows a chi-square distribution with

N - 1

$N-1$ degrees of freedom?)

— Néstor

It would be helpful to modify this answer to include the very strong but unstated assumption that the sample variance follows a chi-squared distribution when the underlying data are independent and follow a normal distribution. Unlike the theory of the distribution of the sample mean, where in practice its sampling distribution will be approximately Normal to reasonable accuracy in many situations, this same asymptotic behavior tends not to happen with the sample variance (until sample sizes become extremely large).

— whuber

Oops. So, so true! This actually came from a problem solution that I handed out to some students, where I state on the question all these assumptions. I edited the answer now.

— Néstor

@user34756 The reason we don't use the distribution of

S^{2}

$S^2$ directly is that its distribution depends on the value of a parameter. You may find it useful to investigate the use of pivotal quantities in constructing confidence intervals.

— Glen_b -Reinstate Monica

Isn't

f (z) = e^{- z^{2} / 2}

$f(z) = e^{-z^2/2}$ instead of

f (z) = e^{- z^{2}}

$f(z) = e^{-z^2}$ ?

— Benoît Legat

বৈকল্পিকের জন্য একটি আস্থা অন্তর তৈরি করার সময় চি স্কোয়ারটি কেন ব্যবহার করা হয়?

Quick answer

Long answer

The proof

Calculating the Confidence Interval for the variance.