Normal Distribution and Dice game

Question

Suppose I use the following rules to generate random numbers. I roll a fair $D_6$ twice. The first roll I call $\mu$ and the second $\sigma^2$. I then Generate a normal random variable with mean $\mu$ and variance $\sigma^2$. I call this $X$

In more concise notation: $X \sim N(U_1,U_2)$ where $U_1,U_2$ are identical independent discrete uniform R.V's on $\{1,2,3,4,5,6\}$

Find an $x \in \mathbb{R} $ such that $\mathbb{P}[X\leq x] = \frac{1}{2}$

I think $x=3.5$.

Reasoning: Symmetry. More precisely: $\mathbb{P}[X \leq x] = \sum\limits_{i=1}^6\mathbb{P}[X\leq x|U_1=i]\cdot\mathbb{P}[U_1=i]=\sum\limits_{i=1}^6\mathbb{P}[X\leq x|U_1=i]\cdot\frac{1}{6}$

Now examine each term in the sum, $\mathbb{P}[X\leq x|U_1=i]\cdot\mathbb{P}[U_1=i]$. Let us take $i=1 $ and $i=6$ as $3.5$ is in the middle of these two values we have that $\mathbb{P}[N(1,U_2)\leq3.5]=\mathbb{P}[N(6,U_2)\geq3.5]$ from symmetry of normal dist. And then finally $\mathbb{P}[N(6,U_2)\leq3.5] = 1- \mathbb{P}[N(6,U_2)\geq3.5] = 1 - \mathbb{P}[N(1,U_2)\leq3.5]$

Hence: $\mathbb{P}[N(1,U_2)\leq3.5]+\mathbb{P}[N(6,U_2)\leq3.5] = \mathbb{P}[N(1,U_2)\leq3.5] + 1- \mathbb{P}[N(1,U_2)\leq3.5]=1 $

So if we split the sum into the $3$ pairs of integers that have $3.5$ in the middle, i.e sum to $7$ i.e $\{(1,6) , (2,5) , (3,4) \}$ we end up with:

$\mathbb{P}[X \leq x] = \sum\limits_{i=1}^6\mathbb{P}[X\leq x|U_1=i]\cdot\frac{1}{6} = \frac{1}{6}(1+1+1)=\frac{1}{2}$

Is this correct? :)

The second part of the question is to now have $X \sim N(U_1^2,U_2)$ where $U_1,U_2$ are identical independent discrete uniform R.V's on $\{1,2,3,4,5,6\}$ Find an $x \in \mathbb{R} $ such that $\mathbb{P}[X\leq x] = \frac{1}{2}$

And then finally to have $X \sim N(U_1^n,U_2)$ for some $n \in \mathbb{N}$ and where $U_1,U_2$ are identical independent discrete uniform R.V's on $\{1,2,3,4,5,6\}$ Find an $x \in \mathbb{R} $ such that $\mathbb{P}[X\leq x] = \frac{1}{2}$

Answer 1

You describe a mixture distribution of six normal distributions $\mathsf{Norm}(\mu=i, \sigma = \sqrt{i}),$ for $i = 1,2,3,4,5,6.$ Perhaps see Wikipedia on mixture distributions, under 'Moments'.

Simulation in R of a million iterations of your procedure, from which we may expect about 2-place accuracy:

set.seed(103)
m = 10^6
u1 = sample(1:6, m, rep=T)
u2 = sample(1:6, m, rep=T)
x = rnorm(m, u1, sqrt(u2))
median(x)
[1] 3.497238  $ aprx 3.5


hist(x, prob=T, col="skyblue2", main="Simulated Mixture Distribution")
 curve(dnorm(x, mean(y), sd(y)), add=T, col="orange", lwd=2)

Note: In this case the mixture distribution is approximately normal. For example, human heights are often modeled as normal, but may actually be mixture distributions of heights of men and women in diverse ethnic groups.

However, in general, mixtures of normal distributions are not (necessarily) anywhere near normal. The 50:50 mixture of $\mathsf{Norm}(100, 10)$ and $\mathsf{Norm}(130, 10)$ is bimodal because the difference in means is more than two standard deviations.

set.seed(2021)
m = 10^6
x1 = rnorm(m, 100, 10);  x2 = rnorm(m, 130, 10)
p = rbinom(m, 1, .5);  y = p*x1 + (1-p)*x2
hist(y, prob=T, col="skyblue2")

Answer 2

Consider the following slightly more general situation. We fix some finite index set $K$, in the OP it is the set of all tuples $(i,j)$, with uniform probability on it, and consider a family $(X_k)$ of independent gaussian random variables, indexed by $k\in K$, so that $X_k\in\mathcal N(\mu_k,\sigma^2_k)$. We consider a two-steps process, in the first step we pick $k$ uniformly, (well, this can be easily generalized using some weights,) in the second step we generate a random number by $X_k$.

(I think, i managed to rewrite equivalently the modelling intention from the OP, the stressed independence is essential below.)

This is equivalent to considering the gaussian random variable $X$ in one step, which models the same experiment, $$ X=\frac 1{|K|}\sum_{k\in K} X_k\ . $$ Its mean is $$ \mu =\Bbb E [X]=\frac 1{|K|} \sum_k\Bbb E[X_k]=\frac 1{|K|} \sum_k \mu_k\ , $$ and the variance is not needed, since we already know that $$ \Bbb P[X\le \mu]=\frac 12\ . $$ For the case "$N(U_1^n,U_2)$" the mean is explicitly $$ \mu=\frac 16(1^n+2^n+3^n+4^n+5^n+6^n)\ . $$ (I may insert some simulation, if this is really a plus.)