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I am going through the book Introduction to Probability by Joseph Blitzstein, Jessica Hwang, and I found the following problem on mixture of two Normal distributions:

A certain stock has low volatility on some days and high volatility on other days. Suppose that the probability of a low volatility day is $p$ and of a high volatility day is $q = 1 -p$, and that on low volatility days the percent change in the stock price is $\mathcal{N}(0, \sigma_1 ^ 2)$, while on high volatility days the percent change is $\mathcal{N}(0, \sigma_2 ^ 2)$, with $\sigma_1 < \sigma_2$.

Let $X$ be the percent change of the stock on a certain day. The distribution is said tobe a mixture of two Normal distributions, and a convenient way to represent $X$ is as $X = I_1 X_1 + I_2 X_2$, where $I_1$ is the indicator random variable of having a low volatility day, $I_2 = 1 - I_1$, $X_j \sim \mathcal{N}(0, \sigma_j ^ 2)$, and $I_1, X_1, X_2$ are independent.

I was trying to write a sample space and then to define random variables $I_1, X, X_1, X_2$ for the above scenario. Let $L$ denote a low volatility day and $H$ denote a high volatility. We can represent the percentage change in stock price by $x$. Therefore we can write sample space as set of all ordered pairs of the form $(T, x)$, where $T \in \{L, H\}$ and $x \in \mathbb{R}$. In other words, the sample space $\Omega$ is given by $$\Omega = \left\{(T, x) : T \in \{L, H\}, x \in \mathbb{R} \right\}$$ I also defined random variables $I_1: \Omega \to \{0, 1\}$, $X: \Omega \to \mathbb{R}$, $X_2$ as follows: $$ I_1((T, x)) := \begin{cases} 1 & \quad \text{if } T = L \\ 0 & \quad \text{if } T = H \end{cases} $$ and $$ X((T, x)) := x.$$ I am comfortable until this. But instead of defining $X$ as above, suppose we try to define $X$ as $X := I_1 X_1 + I_2 X_2$. This means that we first need to define $X_1$ and $X_2$, which I am having trouble with. Conditional on $I_1 = 1$, we have $X_1 \sim \mathcal{N}(0, \sigma_1 ^ 2)$. In other words, if we consider a 'new sample space' $\Omega ^ \prime = \{(L, x) : x \in \mathbb{R}\}$, we can define $X_1 : \Omega ^ \prime \to \mathbb{R}$ by $X_1((L, x)) := x$ and say that $X_1 \sim \mathcal{N}(0, \sigma_1 ^ 2)$. But how do we define $X_1$ on the old sample space $\Omega$? If we tried defining $X_1((T, x))$ as $x$ if $T = L$ and $0$ if $T = H$, then it would be disaster since then $P(X_1 = 0) = 1 - p \neq 0$, which contradicts the fact that $P(X_1 = k) = 0$ for all $k \in \mathbb{R}$. So how can we define $X_1$ and $X_2$, which we can further use to define $X$?

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  • $\begingroup$ You're done--and have done a great job--at the point where you write "I am comfortable with this." For the ensuing question, note that $X=I_1X_1+I_2X_2$ is satisfied by $X_1=X_2=X$ because $$X = 1X = (I_1+I_2)X = I_1X+I_2X.$$ $\endgroup$ – whuber Jun 23 '18 at 11:53
  • $\begingroup$ Correct me if I am wrong. I don't think setting $X_1 = X_2 = X$ would work since then it violates the assumption that $X_1$ and $X_2$ are independent. $\endgroup$ – Supreeth Narasimhaswamy Jun 23 '18 at 12:01
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    $\begingroup$ Independence is a concept related to probability (and cannot be defined without it). You have not (yet) introduced any probability measure. $\endgroup$ – whuber Jun 23 '18 at 12:04
  • $\begingroup$ Suppose we introduce a probability measure $P$ such that $0< P(X_1 \in A) < 1$. This also means that $0< P(X_2 \in A) < 1$ since $X_1 = X_2$. Thus conditional on $X_2 \in A$, the conditional probability of $X_1$ lying in $A$ will be $1$, and therefore $X_1$ and $X_2$ are not independent. What I am trying to say is that if we assign $X_1 = X_2 = X$, does it not make $X_1$ and $X_2$ dependent irrespective of the probability measure we assign? $\endgroup$ – Supreeth Narasimhaswamy Jun 23 '18 at 12:20
  • $\begingroup$ If you want to assure independence, add another dimension. Let, say, the $X_i$ be the projection maps on $\mathbb{R}^2$ to its two coordinates. Your sample space would be $\Omega=\{0,1\}\times \mathbb{R}^2.$ $\endgroup$ – whuber Jun 23 '18 at 12:39
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When thinking about and using random variables, it helps to master some basic constructions. The most fundamental is creating a procedure to produce a random variable with any desired distribution.

Motivation: Why This is Important and What It's Good For

Observe that all computer simulation ultimately employs that most basic function, the uniform random number generator: each time it is called, it produces a number between $0$ and $1$; the numbers it produces from one invocation to the next are supposed to be independent, and the chance that any number lies between limits $0\le a\le b \le 1$ is $b-a.$

Theory proceeds in a similar way. Suppose, then, that you wish to produce a random variable $X$ out of three independent random variables $X_1,$ $X_2,$ and $I$ as in this question. In the question you describe an algebraic combination $X = IX_1 + (1-I)X_2.$ The issue concerns constructing the trivariate random variable

$$(X_1, X_2, I)$$

so that (a) all three components are independent and (b) they have specified probability distributions.

This is done in two steps: (1) forming the sample space $\Omega$ and (2) creating the probability distribution.


Intuition

There are many equivalent ways to proceed. Before doing so, permit me to share some intuition, which I draw from the tickets in a box model. The "tickets" (slips of paper) in a box are the sample space $\Omega;$ what you write on them are values of random variables. To produce independent variables, first create a box representing one variable, writing its values on each ticket and creating enough tickets of each value to give the intended probability.

Next, pull each of those tickets systematically out of the box. Use each one to create a new box of tickets in which two numbers are written: the first is the value on the pulled ticket and the second is a possible value of the second random variable. This new box gives values of the second random variable "conditioned on" the first. The proportions of each new ticket give the conditional distribution.

If you make the proportions always the same, so that the conditional distribution does not vary, you will create two independent random variables simply by taking all the new ("conditional") boxes you created and dumping their contents into one big box. The values written on the tickets in the big box are all ordered pairs of numbers: one for the first random variable and another for the second.

What we need, then, in our construction are (1) a mathematical way to create ordered pairs--that's obviously the Cartesian product--and (2) a formal way to specify the proportions. That's given by the distribution functions $F$ of the variables, or rather by their inverses: given any proportion $q$, the value of the inverse $x=F^{-1}(q)$ tells you what fraction of tickets must have values of $x$ or less written on them.


Preliminary Definitions and Notation

Let's take as our point of departure these quantile functions for the random variables. To be specific, suppose you are given a distribution function $F$. By definition, $F:\mathbb R \to [0,1].$ Its quantile function $X_F$ (also written $F^{-1}$) is defined on the interval $\mathcal{I}=[0,1]$ and takes values in the "extended real line" $$\mathbb{R}_\omega = \{-\infty\} \cup \mathbb{R} \cup \{\infty\}$$ according to the rule

$$X_F(q) = \sup\,\{x\in \mathbb{R}_\omega\mid F(x) \lt q.\}.$$

This has the essential property of mapping each subinterval $[0,q]$ into the set of all numbers $x$ for which $F(x)\le q$:

$$\{x \in \mathcal{I}\mid X_F(x) \le q\} = [0, q].\tag{*}$$

The Sample Space

Given a tuple of distribution functions $F_1, \ldots, F_n$ define

$$\Omega = [0,1]^n$$

to be the Cartesian product of the unit interval.

The Probability Measure

The probability measure on $\Omega$ is the uniform one: that is, for all $0\le a_i\le b_i\le 1$ with $i=1,2,3,\ldots, n,$

$$\Pr\left(\omega \in [a_1,b_1] \times [a_2,b_2] \times \cdots \times [a_n,b_n]\right) = (b_1-a_1)(b_2-a_2)\cdots(b_n-a_n).\tag{**}$$

The Construction

For $\omega=(q_1,q_2,\ldots,q_n)\in\Omega,$ define

$$X_i(\omega) = X_{F_i}(q_i)$$

to be the quantile function of $F_i$ evaluated at the $i^\text{th}$ component of $\omega.$ The tuple $(X_1,X_2,\ldots,X_n)$ is the desired set of independent random variables on $\Omega$ having the given (marginal) distributions.

(The fact that they are random variables includes an assertion that they are measurable functions--but because I wish to focus on the statistical properties of the variables, I will omit all discussion of the measure-theoretic aspects of this construction and leave the corresponding parts of the proofs as a (simple) exercise for those who are interested.)

Proofs

We have to demonstrate two things: that the $X_i$ are independent and that each $X_i$ has $F_i$ for its distribution. Together, these mean that for all extended real numbers $x_1, x_2, \ldots, x_n,$ the distributional values are given by the product of the $F_i$:

$$\Pr(X_1 \le x_1, X_2 \le x_2, \ldots, X_n \le x_n) = F_1(x_1) F_2(x_2) \cdots F_n(x_n).$$

It is almost trivial that this is so, because by virtue of $(*)$ above,

$$\{\omega\in\Omega\mid X_1(\omega)\le x_1, \ldots, X_n(\omega)\le x_n\} = [0, F_1(x_1)] \times[0, F_2(x_2)] \times \cdots \times [0, F_n(x_n)]$$

which under the uniform distribution $({**})$ has probability $$(F_1(x_1)-0)(F_2(x_2)-0)\cdots (F_n(x_n)-0) = F_1(x_1)F_2(x_2)\cdots F_n(x_n),$$ QED.

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Your description before the sentence "I am comfortable until this" does not make mention of a probability measure (in particular, $p$ and the variances do not occur there). But in particular the measure on the real axis that you did not specify here is not "simple"; it is a mixture distribution. Mixture distributions are easiest described by a two-step process using a probabilistic interpretation with conditional distributions, as described. The "direct" probability model that you seek would require specifying a two-mode Gaussian mixture, which is more tedious, but not necessary for simulations.

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  • $\begingroup$ Can this explained without using measure theory or advanced probability? This is my first course on Probability, and I do not know Measure Theory yet. $\endgroup$ – Supreeth Narasimhaswamy Jun 23 '18 at 9:40
  • $\begingroup$ My own comment, while not going into as much depth as @whuber, only used the word "measure", which is perfectly standard in non-measure theoretic probability, at least if you replace it by the synonymous "probability". Basically, if you want to get to the theoretical bottom of things, then measure theory is really impossible to avoid. $\endgroup$ – Christian Rau Jun 24 '18 at 10:33

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