A sample of normal RVs - the distribution of Xi-Xbar?

[Phillips101]

We have X1,...,Xn~N(mu, sigma2)

The crux of my problem is finding out the distribution of, say, X1-Xbar (where Xbar is the mean of the n RVs). This is going to end up proving the independence of Xbar and Sxx, btw.

I know Xbar~N(mu, sigma2/n), but I don't know how to find the distribution of a difference of normal RVs with different arguments?

Thanks for any help.

 

[statdad]

Notice that

$$X_1 - \bar X = (1-\frac 1 n) X_1 - \frac 1 n \sum_{i\ge2}X_i$$

and all of $X_1$ and $X_2, \dots, X_n$ are independent.

* Get the distribution of

$$(1 - \frac 1 n) X_1$$

as well as that of

$$\frac 1 n \sum_{n\ge2} X_i$$

These are independent as well, so find the distribution of their difference.

 

[Phillips101]

I have Xi-Xbar ~ N(0, (1+1/n)sigma2) ?

 

[statdad]

Are you sure the variance is

$$\left(1 + \frac 1 n \right) \sigma^2$$

 

[blue_raver22]

As a scientist, it is important to understand the distribution of the difference between two random variables. In this case, we are interested in the distribution of X1-Xbar, where Xbar is the mean of n RVs. This is a common problem in statistics and can be solved using the central limit theorem.

The central limit theorem states that when the sample size is large enough, the distribution of the sample mean will approach a normal distribution, regardless of the underlying distribution of the individual RVs. In this case, we have n RVs that are normally distributed with mean mu and variance sigma2. Therefore, by the central limit theorem, Xbar will also be normally distributed with mean mu and variance sigma2/n.

Now, let's consider the distribution of X1-Xbar. This can be rewritten as (X1-Xbar)-(mu-mu). Using basic algebra, we can see that this is equivalent to (X1-mu)-(Xbar-mu). We know that X1-mu is normally distributed with mean 0 and variance sigma2, and Xbar-mu is normally distributed with mean 0 and variance sigma2/n. Therefore, by the properties of normal distributions, the difference (X1-Xbar) will also be normally distributed with mean 0 and variance sigma2+sigma2/n.

In summary, the distribution of X1-Xbar will be a normal distribution with mean 0 and variance sigma2+sigma2/n. This result also proves the independence of Xbar and Sxx, as their distributions are both normal and their means and variances are not affected by each other. I hope this helps in understanding the distribution of X1-Xbar.