Let  $X_1, X_2, ..., X_n$ be normally distributed with mean $\mu$ and variance $\sigma ^2$, then $\frac{(n-1) S^2}{ \sigma^2}$ has a Chi-Square distribution with $n-1$ degrees of freedom.   How come it's chi-square distributed?  Attempt:  $S^2 = \frac{1}{n-1} \sum(X_i-\overline X)^2 = \frac{1}{n-1}(\sum X_i ^2 - (\sum X_i)^2)$.    Here, the first term $\sum X_i^2$ is chi-squared. But you also need to subtract the second term (which is normal distributed with mean $\mu$, and variance $\frac{\sigma^2}{n}$. How does it make it chi-square distributed?
36151
It's a standard result that given $X_1,\cdots ,X_n $ random sample from $N(\mu,\sigma^2)$, the random variable $$\frac{(n-1)S^2}{\sigma^2}$$ has a chi-square distribution with $(n-1)$ degrees of freedom, where $$S^2=\frac{1}{n-1}\sum^{n}_{i=1}(X_i-\bar{X})^2.$$ I would like help in proving the above result. Thanks.
76136
How can I prove that the sum of $X_1, X_2, \ldots,X_n$ random variables, all of which have normal distributions $N(\mu_i, \sigma_i)$, is a random variable that is itself normally distributed with mean $$\mu =\sum_{i=1}^n \mu_i$$  and variance  $$\sigma^2 = \sum_{i=1}^n \sigma_i^2$$  Edit: I forgot to add that this was with the assumption that all $X_1, X_2,\ldots,X_n$ are independent.
51996
I've been trying to establish that the sample mean and the sample variance are independent. One motivation is to try and write the sample variance, $S^{2}$ as a function of $\left\{ X_{2}-\bar{X},X_{3}-\bar{X},\cdots,X_{n}-\bar{X}\right\} =A$ only. Then we proceed by showing that $A$ and $\bar{X}$ are independent (which I'm unable to show), which then implies the independence of $S^{2}$ and $\bar{X}.$    I would appreciate it if the good people of M.SE would help guide me in the right direction. Thanks.  Edit: The random samples $X_1,\cdots,X_n$ are from an $N(\mu, \sigma)$ distribution.
113628
I am trying to understand the logic behind chi-squared test.  The Chi-squared test is $\chi ^2 = \sum \frac{(obs-exp)^2}{exp}$. $\chi ^2$ is then compared to a Chi-squared distribution to find out a p.value in order to reject or not the null hypothesis. $H_0$: the observations come from the distribution we used to created our expected values. For example, we could test if the probability of obtaining head is given by $p$ as we expect. So we flip 100 times and find $n_H$ Heads and $1-n_H$ tails. We want to compare our finding to what is expected ($100 \cdot p$). We could as well use a binomial distribution but it is not the point of the question… The question is:   Can you please explain why, under the null hypothesis, $\sum \frac{(obs-exp)^2}{exp}$ follows a chi-squared distribution?  All I know about the Chi-squared distribution is that the chi-squared distribution of degree $k$ is the sum of $k$ squared standard normal distribution.
120952
Suppose $X=[X_{1},X_{2}]$ and $X$~$N_2(μ,Σ)$. I wish to find the distribution of  $5X_{1}^2+2X_{1}X_{2}+X_{2}^2$. Since this is of a quadratic form I do not know a way of solving this. However I kind of feel like its chi-squared distributed with parameter 2. Any answers will be welcomed. Thanks
97436
If I have the expression below, how can I use y~N($\mu$, $\Sigma$) to show it is a $\chi^2$ distribution with ($_n-1$) degrees of freedom?  ($n$-1)($s^2$)/$\sigma^2$ how is this ~$\chi^2$ $_n$-$_1$?     I recognize this contains SSE. I know that I should show the square root produces a distribution of $N$($\mu$, $I$) . I am lost at this point. I am thinking perhaps to substitute quadratic form into the original expression, then see what cancels? Or is this totally wrong?
88502
Question  Suppose we have a multinomial distribution with $N$ possible outcomes, with probabilities $p_1,\ldots,p_N$. We sample this $n$ times, and denote the observed frequency of the $i$th outcome as $\xi_i$. In [1] the author claims that the distribution of the $\xi_i$ in the limit of large $n$ is:  $$p(\xi_1,\ldots,\xi_N)\propto\exp\left(-\frac{n}{2}\sum_{i=1}^N\frac{(\xi_i-p_i)^2}{p_i}\right).\;\;\;\;\;(1)$$  We can see immediately that this must be an approximation, as it assigns nonzero probabilities for $\xi_1+\cdots+\xi_N&gt;1$. However we can see that these have vanishing probability in the limit $n\rightarrow\infty$. My question is how do we derive (1) from the multinomial distribution, and show that they match in the $n\rightarrow\infty$ limit?  My thoughts  My first thought would be to appeal to the central limit theorem. The multinomial distribution has mean $\mu_i=p_i$ and covariance matrix $\Sigma_{ij}=\delta_{ij}p_i-p_ip_j$, so we would expect this in the large $n$ limit to be described by a multivariate Gaussian with mean $\mu$ and covariance $\frac{1}{n}\Sigma$. However, things are complicated by the fact that the multinomial covariance is singular (since $\xi_N$ is determined by the other $\xi_i$s), and so the multivariate Gaussian is not defined.  To address this, we may try and consider only the first $\xi_1,\ldots,\xi_{N-1}$, which have a non-singular covariance matrix and hence well-defined multivariate Gaussian distribution. Let's take the Binomial distribution $N=2$. The frequency $\xi_1$, this has mean $p_1$ and variance $p_1(1-p_1)$, so this would be described the the Gaussian: $$\propto\exp\left(-\frac{n}{2}\frac{(\xi_1-p_1)^2}{p_1(1-p_1)}\right).\;\;\;\;\;(2)$$ The expression (1) gives: $$\propto\exp\left(-\frac{n}{2}\left(\frac{(\xi_1-p_1)^2}{p_1}+\frac{(\xi_2-p_2)^2}{p_2}\right)\right).\;\;\;\;\;(3)$$ If we substitute $\xi_2\rightarrow 1-\xi_1$, $p_2\rightarrow 1-p_1$ into (3), we can verify that this gives the same answer as (2). I have verified that this also works for $N=4$.   I'm sure that if I just bashed out the algebra for general $N$ we would get agreement between the central limit theorem and (1) when we restrict the latter to $\xi_1+\cdots+\xi_N=1,p_1+\cdots+p_N=1$. However, how can we start with the multinomial distribution and derive (1) as a limit which is valid everywhere? One idea would be to say that (1) goes to zero as $n\rightarrow\infty$ when you are not on that plane, however I am a bit uncomfortable with this as it goes to zero everywhere except the mean as $n\rightarrow\infty$, so I don't know if that argument is good enough.  [1] Wootters, William K. "Statistical distance and Hilbert space." Physical Review D 23.2 (1981): 357.
85658
Let random variables $X$ and $Y$ be independent Normal with distributions $N(\mu_{1},\sigma_{1}^2)$ and $N(\mu_{2},\sigma_{2}^{2})$. Show that the distribution of $(X,X+Y)$ is bivariate Normal with mean vector $(\mu_{1},\mu_{1}+\mu_{2})$ and covariance matrix  $$ \left( \begin{array}{ccc} \sigma_{1}^2 &amp; \sigma_{1}^2  \\ \sigma_{1}^2 &amp;\sigma_{1}^2+\sigma_{2}^2  \\  \end{array} \right).$$  Thanks .
57761
Suppose that $\mathbf{X} \sim N_{2}(\mathbf{\mu}, \mathbf{\Sigma})$. Then the conditional distribution of $X_1$ given that $X_2 = x_2$ is multivariate normally distributed with mean:   $$ E[P(X_1 | X_2 = x_2)] = \mu_1+\frac{\sigma_{12}}{\sigma_{22}}(x_2-\mu_2)$$  and variance: $${\rm Var}[P(X_1 | X_2 = x_2)] = \sigma_{11}-\frac{\sigma_{12}^{2}}{\sigma_{22}}$$  It makes sense that the variance would decrease since we have more information. But what is the intuition behind the mean formula? How does the covariance between $X_1$ and $X_2$ factor into the conditional mean?
21152
I am trying to find distribution of random variable:  $$\frac{X_1+X_2 X_3}{\sqrt{1+X_3^2}}$$  where $X_i \sim \mathcal{N}(0,1)$ and independent.  My thinking so far is to use random variable algebra and to find distributions step by step. However, I don't know what to do with the variable: $\sqrt{1+X_3^2}$ (it seems like some variant of $\chi$-distribution).   Moreover, I'm worried that my thinking is to complicated and that the solution should be more elegant.
62388
How can we derive the chi-squared probability density function (pdf) using the pdf of normal distribution?  I mean, I need to show that   $$f(x)=\frac{1}{2^{r/2}\Gamma(r/2)}x^{r/2-1}e^{-x/2} \&gt;, \qquad x &gt; 0\&gt;.$$
19793
The statement  The sampling distribution of the sample variance is a chi-squared distribution with degree of freedom equals to $n-1$, where $n$ is the sample size (given that the random variable of interest is normally distributed).    My intuition  It kinda makes intuitive sense to me 1) because a chi-square test looks like a sum of square and 2) because a Chi-squared distribution is just a sum of squared normal distribution. But still, I don't have a good understanding of it.  Question  Is the statement true? Why?
94387
Student's t distribution is defined as the ratio of a standard normally distributed random variable and the square root of a Chi-square distributed random variable divided by its degrees of freedom, given that they are independent. In formulas one can write $\frac{Z}{\sqrt \frac{U}{df}}$, where $Z$ is $N(0,1)$ and $U$ is $\chi^2_{df}$.   In showing that this statement is true, I arrived at the point in which I have $\frac{(n-1)S^2}{\sigma^2}\sim\chi^2_{n-1}$ and $\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt n}}\sim N(0,1)$. Then, following the definition, we would have that  \begin{gather} \frac{\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt n}}}{\sqrt\frac{\frac{(n-1)S^2}{\sigma^2}}{n-1}} \end{gather}  is distributed as a $t_{n-1}$. But I am stuck at how to prove than this two random variables are independent between them.We covered a result about independence in the case of two Chi-square random variables and I thought of seeing the standard Normal as the square of a Chi-square random variable but I am afraid of it being mathematically sacrilegious.  Do you have any hint?
37957
I want to solve the following problem:  Suppose that $X$ and $Y$ are two independent, Gaussian variables, with mean zero and common variance $\sigma^2$. Let $R^2 = X^2+ Y^2$ and $Z = \arctan (X/Y)$. I know that in this case $R^2$ has an exponential distribution and $Z$ has a uniform distribution. However, are $R$ and $Z$ independent from each other?
22712
... and why ?  Assuming $X_1$,$X_2$ are independent random-variables with mean $\mu_1,\mu_2$ and variance $\sigma^2_1,\sigma^2_2$ respectively. My basic statistics book tells me that the distribution of the $X_1-X_2$ has the following properties:    $E(X_1-X_2)=\mu_1-\mu_2$ $Var(X_1-X_2)=\sigma^2_1 +\sigma^2_2$   Now let's say $X_1$, $X_2$  are t-distributions with $n_1-1$, $n_2-2$ degrees of freedom. What is the distribution of $X_1-X_2$ ?   This question has been edited: The original question was "What are the degrees of freedom of the difference of two t-distributions ?". mpiktas has already pointed out that this makes no sense since $X_1-X_2$ is not t-distributed, no matter how approximately normal $X_1,X_2$ (i.e. high df) may be.
101580
Ok, so the Chi-Squared distribution with $n$ degrees of freedom is the sum of the squares of $n$ independent Gaussian random variables.  The trouble is, my Gaussian random variables are not independent. They do however all have zero mean and the same variance. Supposing I have a covariance matrix---which again is not a diagonal matrix because they aren't independent, but all the elements along the diagonal are equal to each other because they have the same variance, and in fact the covariance matrix is a symmetric Toeplitz matrix (and I'm not saying that this is important to the solution if there is one, but if it's a necessary property to get anywhere, by all means use that fact)---is there some way to decompose this sum of squares of these Gaussian random variables into perhaps a sum of chi-squared random variables and possibly Gaussian random variables? In other words, I can't directly just square them all and add them together and call it a chi squared distribution because a chi squared distribution is a sum of independent Gaussian squares, and they aren't independent.  I know how to find a linear transformation of the Gaussian random variables which are $n$ independent Gaussians, but that's no help because they aren't the things being squared, you see.
112321
Let $X_1, \ldots, X_n$ be i.i.d. exponentially distributed random variables with  density  $$\eqalign{\theta^{-1} e^{-x/\theta}, &amp;x \ge 0 \\ 0, &amp;x \lt 0} $$  and let $Y_i = X_{(i)}$ denote the order statistics such that $Y_1 \leq \cdots \leq Y_n$.  How to show that $$     2\frac{\left(\sum_{i=1}^{r}Y_i\right) + (n-r)Y_r}{\theta} $$     has a chi-square distribution with $2r$ degrees of freedom?  I wrote the joint density of $(Y_1,Y_2,...,Y_r)$ but nothing became apparent.