Distribution of the ratio of two Normal variables

Question

Considering a sample of size $n$ from a $N(\mu,\sigma^2)$ distribution: $X_1, \ldots , X_n$, I need to find the ratio of \begin{equation} R = \frac{\tilde{\mu} - \mu}{\hat{\mu} -\mu}, \end{equation} where $\tilde{\mu} = X_1 $ one observation, and $\hat{\mu} = \bar{X}$, the sample mean.

I know that $ \frac{\tilde{\mu} - \mu}{\sigma} \sim N(0,1)$, $ \frac{\sqrt{n}(\hat{\mu} - \mu)}{\sigma} \sim N(0,1)$, and the ratio of two normally distributed variables is Cauchy - for this to happen do we need for the variables to be independent? Also, there is a $\sqrt{n}$ coming in the picture when simplifying the ratio that I am not sure how to handle.

Answer 1

For real $\alpha$ and $\beta>0$, suppose $\text{Cauchy}(\alpha,\beta)$ denotes the density $f(x)=\frac{\beta}{\pi((x-\alpha)^2+\beta^2)}\,,x\in \mathbb R$.

It can be shown using a change of variables or otherwise that if $(X,Y)$ has a standard bivariate normal distribution with zero means, unit variances and correlation $\rho$, then $\frac{X}{Y}$ has a $\text{Cauchy}(\rho,\sqrt{1-\rho^2})$ distribution. Wikipedia states a more general result which agrees with this.

It is clear that the distribution of $R$ is free of $\mu,\sigma$ because $$R=\frac{X_1-\mu}{\overline X-\mu}=\frac{(X_1-\mu)/\sigma}{(\overline X-\mu)/\sigma}=\frac{Y_1}{\overline Y}\,,$$

where $Y_i=(X_i-\mu)/\sigma$ are i.i.d standard normal for all $i=1,\ldots,n$.

We can rewrite this as $$R=\sqrt n\left(\frac{Y_1}{\sqrt n \overline Y}\right)$$

Notice that

\begin{align} \operatorname{Cov}(Y_1,\sqrt n\overline Y)&=\operatorname{Cov}\left(Y_1,\frac1{\sqrt n}\sum_{i=1}^n Y_i\right)& \\&=\frac1{\sqrt n}\sum_{i=1}^n\operatorname{Cov}(Y_1,Y_i) \\&=\frac1{\sqrt n}\operatorname{Var}(Y_1)=\frac1{\sqrt n} \end{align}

Now as $\overline Y$ is a linear combination of independent normal variables, $(Y_1,\sqrt n\overline Y)$ is bivariate normal with zero means, unit variances and correlation $\frac1{\sqrt n}$. The ratio $\frac{Y_1}{\sqrt n \overline Y}$ therefore has a $\text{Cauchy}\left(\frac1{\sqrt n},\sqrt{\frac{n-1}n}\right)$ distribution, from which it follows that

$$R\sim \text{Cauchy}(1,\sqrt{n-1})$$

Answer 2

Since $Y_j:=X_j-\mu\sim N(0,\,\sigma^2)$ are independent, $R=\frac{n}{1+W}$ with $W:=\frac{\sum_{j=2}^nY_j}{Y_1}$ the ratio of independent variables of respective distribution $\sigma\sqrt{n-1}N(0,\,1),\,\sigma N(0,\,1)$. Since $\frac{W}{\sqrt{n-1}}$ has the standard Cauchy distribution, $W$ has CDF $\tfrac12+\tfrac{1}{\pi}\arctan\tfrac{w}{\sqrt{n-1}}$. I'll leave converting this into $R$'s CDF as an exercise.

Answer 3

First notice that $X_1-\mu, \ldots, X_n-\mu \sim\text{i.i.d.} \operatorname N(0,\sigma^2)$ and the distributions of $\widehat\mu-\mu$ and $\widetilde\mu-\mu$ are those of the sample mean and first observation from an i.i.d. sample from $\operatorname N(0,\sigma^2),$ so no generality is lost by assuming $\mu=0.$ Nor is any generality lost by assuming $\sigma=1.$

You seek the distribution of $$ \frac{n X_1}{X_1 + \cdots + X_n} = \frac{\sqrt n}{\sqrt n + \left( \frac{X_2+\cdots+X_n}{X_1\sqrt n} \right)} $$ The expression in $\Big($parentheses$\Big)$ is a quotient of two normally distributed random variables both of which have expected value $0$ and equal variances, and are independent.

As noted in the question that expression therefore has a Cauchy distribution.

Now the question is if $W$ has a Cauchy distribution, then what is the distribution of $\dfrac{\sqrt n}{\sqrt n + W} \text{ ?}$

Here the following becomes relevant: if $W$ is Cauchy-distributed, then so is $\dfrac{aW+b}{cW+d}$ if $ad-bc\ne0,$ although the location and scale parameters may change.

Maybe a proof of that is worth a separate question, or maybe it's already here.