# What does it mean to say that $X_1, X_2$ have a “common” Normal distribution?

Let $$X_1, X_2$$ be rvs having a common Normal distribution $$N(0,1)$$ with $$\operatorname{Corr}(X_1, X_2) = \rho$$. Calculate the coefficient of upper tail-dependence for all $$\rho \in [-1, 1]$$.

What does it mean with it says they have a "common" Normal distribution?

My first thought was that they meant both $$X_1$$ and $$X_2$$ are univariate normal $$N(0,1)$$ distributed variables. However, if that is true, then the question doesn't make sense. The tail-dependence cannot be calculated.

So I am left to believe that by "common" Normal distribution, they mean the bivariate Normal distribution?

It means that two things are true.

First:

$$P(X_1 < t) = P(X_2 < t)$$

for all real numbers $$t$$ (i.e., $$X_1$$ and $$X_2$$ have the same distribution, often the shorthand equidistributed is used to describe this condition).

Second:

$$P(X_1 < t) = \frac{1}{\sigma \sqrt{2 \pi}}\int_{-\infty}^t e^{\frac{(x - \mu)^2}{2 \sigma^2}} \,\text{d}x$$

for some fixed numbers $$\mu$$ and $$\sigma$$ (i.e. the distribution of $$X_1$$ (*) is a normal distribution).

This doesn't imply that $$(X_1, X_2)$$ is joint normal without further assumptions. If that was intended, it's not what the author actually wrote.

(*) Given the first condition, this implies that the distribution of $$X_2$$ is also a normal distribution.

I think "common" here just means that the marginal distribution $$\text{N}(0,1)$$ is common to both random variables (i.e., they have the same marginal distribution). Although technically this is insufficient to give a bivariate normal distribution, I think the writer probably intended that form:

$$\begin{bmatrix} X \\ Y \end{bmatrix} \sim \text{N} \Big( \begin{bmatrix} 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 & \rho \\ \rho & 1 \end{bmatrix} \Big).$$

That specification would yield common marginal distributions $$X \sim \text{N}(0,1)$$ and $$Y \sim \text{N}(0,1)$$. If I were you, I would suggest noting this technicality, and then proceed on the basis that the random variables are bivariate normal. You might want to note the issue again as a caveat once you give your answer.

The exercise is badly phrased. I suspect what is meant is that the two random variables are jointly normal and have a common distribution. If they're separately normal but not jointly normal, then you don't have enough information to answer the question. If my suspicion is right, then the exercise should have said they are jointly normal.

To have a "common" distribution simply means they both have the same distribution. Thus: \begin{align} \require{cancel} & \left[ \begin{array}{l} X_1 \\ X_2 \end{array} \right] \sim \xcancel{\operatorname N\left( \left[ \begin{array}{l} \mu_1 \\ \mu_2 \end{array} \right], \left[ \begin{array}{cc} \sigma_1^2 & \rho \sigma_1\sigma_2 \\ \rho\sigma_1\sigma_2 & \sigma_2^2 \end{array} \right] \right)} & & \longleftarrow \text{ not common} \\[10pt] & \left[ \begin{array}{l} X_1 \\ X_2 \end{array} \right] \sim \operatorname N\left( \left[ \begin{array}{l} \mu \\ \mu \end{array} \right], \left[ \begin{array}{cc} \sigma^2 & \rho \sigma^2 \\ \rho\sigma^2 & \sigma^2 \end{array} \right] \right) & & \longleftarrow\text{ common} \end{align} In the second case we have $$X_i\sim\operatorname N(\mu,\sigma^2)$$ for $$i=1,2,$$ thus each is normally distributed and they have that normal distribution in common.