Difference between iid $X,Y \sim F(\cdot)$ and $X=Y$

Question

I'm having a confusion with understanding the equality of rvs.

If $X \sim Exp(1)$ and $Z=2X$, then $Z \sim Exp(\frac{\lambda}{2})$. If also $Y \sim Exp(1)$, then $X$ and $Y$ are iid and $Z=X+Y, Z \sim Erlang(2,1)$. I don't quite understand the case when, for example, $Z=X+X$, or (equivalently?), $Z=X+Y$ when $Y=X$, because I don't fully understand what it means for two rvs to be $equal$.

From the axiomatic point of view, if $\exists$ probability triple $(\Omega, \mathcal{F}, P)$ with two measurable functions $X,Y$ defined on this triple, for these two measurable functions ('random' 'variables') to be equal is to have $X(\omega)=Y(\omega) \ \forall \omega: X,Y \subset \mathbb{R}$ except maybe $\omega$ with Lebesgue measure $0$. The strict inequality, e.g. $X_n(\omega)<X(\omega) \ \forall \ n$ is then used for proving sure/almost sure convergence of rvs.

I don't quite understand how to apply the equality in axiomatic sense to the problem above, e.g. finding the distribution of $Z=X+X$ and $Z=X+Y$.

Answer 1

This only addresses the "...I don't fully understand what it means for two rvs to be equal" part of your question.

Random variables are functions.

• Two random variables $Y_1:(\Omega,\mathscr{F},P)\rightarrow\mathbb{R}$ and $Y_2:(\Omega,\mathscr{F},P)\rightarrow\mathbb{R}$ are equal od $Y_1(\omega)=Y_2(\omega)$ for all $\omega\in\Omega$. In shuch case, the also have the same distribution.

• It could be that two random variables are define in the same probability space, they are not equal and yet, they have the same distribution: $$X_1:((0,1),\mathscr{B}(0,1),\lambda)\rightarrow\mathbb{R}$$ defined by $$X_1(t)=\mathbb{1}_{(0,1/2]}(t)$$ and $$X_2:((0,1),\mathscr{B}(0,1),\lambda)\rightarrow\mathbb{R}$$ defined by $$X_2(t)=\mathbb{1}_{(1/2,1)}$$ where $\lambda$ is the probability measure on $(0,1)$ that assigns to any subinterval $(a,b]\subset(0,1)$ it length: $\lambda((a,b])=b-a$.

Clearly $X_1\neq X_2$. However, since
$$\begin{align} P[X_1=1]&=\lambda(\{t: X_1(t)=1\})=\lambda((0,1/2])=1/2\\ P[X_2=1]&=\lambda(\{t: X_2(t)=1\})=\lambda((1/2,1))=1/2 \end{align} $$ and both $X_1$ and $X_2$ take values $0$ or $1$, we have that $X_1\stackrel{law}{=}X_2$, that is, they have the same distribution.

• Random variables may also be defined in different probability spaces and yet have the same distribution:

$$Z_1:((0,1)\times(0,1),\mathscr{B}((0,1)\times(0,1)),\lambda_2)\rightarrow\mathbb{R}$$ defined by $$Z_1(s,t)=s$$ and $$Z_2:((0,1),\mathscr{B}((0,1)),\lambda)\rightarrow\mathbb{R}$$ defined by $$Z_2(x)=x$$

where $\lambda_2$ is the measure that assigns to each box in $(0,1)\times(0,1)$ it area.

$Z_1$ and $Z_2$ are random variables defined in different probability spaces; but $$\lambda_2(a<Z_1\leq b)=\lambda_2((a,b]\times(0,1))=b-a=\lambda_2(a<Z_1\leq b)$$ Thus, $Z_1\stackrel{law}{=}Z_2$.

Answer 2

To somehow answer your question in comments and give you some insight.

If $X,Y$ are equal almost surely (so that $X=Y$ a.s) then $X,Y$ have the same distribution.

However, if $X,Y$ have the same distribution, it doesn't really mean that $X,Y$ are the same almost surely. In fact in many cases they aren't, because what would be the point of defining another almost equal random variable, already having one.

To stick to your example for a while. Let $X,Y,W \sim \mathcal Exp(\lambda)$ be three random variables, each has Exponential distribution with parameter $\lambda > 0$. Now, let's say $X=Y$ almost surely, whereas $X,W$ are independent. Then random variable $Z=X+Y$ is really different from random variable $V=X+W$, even though every of $X,Y,W$ has the same distribution.

Indeed, $Z=X+Y=2X$ (because $Y=X$ was defined this way), so that $F_Z(t) = \mathbb P(X \le \frac{t}{2}) = \begin{cases} 1- e^{-\frac{\lambda t}{2}} & t \ge 0 \\ 0 & t<0 \end{cases}$ which means $Z = X+Y \sim \mathcal Exp(\frac{\lambda}{2})$.

However, dealing with $V=X+W$, we can't write $V=2X$. To find the distribution of such sum, we need to take a look at distribution of VECTOR $(X,W)$ (in fact in case of $Z$ we should do the same, but $(X,X)$ is specified by $X$, so it was enough to look at $X$). How to find the distribution of $(X,W)$ ? Well, we have an information that these variables are independent. Which means the distribution $\mu_{(X,W)}$ of that vector is in product form $\mu_X \otimes \mu_W$ and in case both has densities, that product distribution has density, too, given by $f(x,w) = \lambda e^{-\lambda x} \lambda e^{-\lambda w} 1_{(0,\infty)^2}(x,w)$ (it's the product of densities of $X,W$)

And we need to calculate (for example CDF)

$F_V(t) = 0$ for $t<0$ (since both $X,W$ are non-negative). Taking $t \ge 0$ we get: $$ F_V(t) = \mathbb P(X+W \le t) = \mathbb P( (X,W) \in A_t)$$ where $A_t = \{(x,w) \in \mathbb R_+^2 : x+w \le t \}$ (note that we now have a vector under probability, so we can work with our distribution of vector, which we know).

$$ F_V(t) = \int_{A_t} f(x,w) dxdw = \lambda^2 \int_0^t e^{-\lambda w} \int_0^{t-w} e^{-\lambda x} dx dw = \lambda \int_0^t e^{-\lambda w} (1-e^{-\lambda(t-w)})dw = \lambda \int_0^t e^{-\lambda w} - e^{-\lambda t} dw = (1-e^{-\lambda t}) -t\lambda e^{-\lambda t} $$ And you can differentiate to find density of $V$ to be:

$$g_V(t) = \begin{cases} 0 & t \le 0 \\ \lambda e^{-\lambda t} - \lambda e^{-\lambda t} + \lambda^2 t e^{-\lambda t} = \lambda^2 t e^{-\lambda t} = \frac{t^{2-1}\lambda^2 e^{-\lambda t}}{(2-1)!} & t>0 \end{cases} $$ Which you might know to be $\Gamma(2,\lambda)$ or sometimes it is called $Erlang(2,\lambda)$ or $\Gamma(2,\frac{1}{\lambda})$