# Write some probabilities using a common cumulative distribution function

Consider the random variables $Y,\epsilon_1, \epsilon_2$. $\epsilon\equiv (\epsilon_1, \epsilon_2)$ has cumulative distribution function (cdf) $F$.

Let $u_1, u_2$ be some known real-valued parameters.

Suppose $Y$ can take values $\{0,1,2\}$ with the following probabilities $$(\diamond) \text{ } \begin{cases} \mathbb{P}(Y=1)=\mathbb{P}(\epsilon_1\geq -u_1, \epsilon_1-\epsilon_2\geq u_2-u_1)\\ \mathbb{P}(Y=2)=\mathbb{P}(\epsilon_2\geq -u_2, \epsilon_1-\epsilon_2\leq u_2-u_1)\\ \mathbb{P}(Y=0)=\mathbb{P}( \epsilon_1\leq -u_1, \epsilon_2\leq-u_2)\\ \end{cases}$$ (I haven't done a precise distinction between weak and strict inequalities as I assume that $F$ is continuous)

To understand better I have pictured an example with $u_1\equiv 3$ and $u_2\equiv -1$. $\mathbb{P}(Y=1)$ is the probability of the pink area, $\mathbb{P}(Y=2)$ is the probability of the green area, $\mathbb{P}(Y=0)$ is the probability of the blue area.

Question: I want to re-write these probabilities using one joint cdf only (not necessarily $F$). Is there a way to do that?

My thoughts:

The tricky part is that

• these probabilities depend on different random variables: $\mathbb{P}(Y=1)$ on $\epsilon_1, \epsilon_1-\epsilon_2$; $\mathbb{P}(Y=2)$ on $\epsilon_2, \epsilon_1-\epsilon_2$; $\mathbb{P}(Y=0)$ on $\epsilon_1, \epsilon_2$

• the cdf's of $\begin{pmatrix} \epsilon_1\\ \epsilon_1-\epsilon_2 \end{pmatrix}$, $\begin{pmatrix} \epsilon_2\\ \epsilon_1-\epsilon_2 \end{pmatrix}$, $\begin{pmatrix} \epsilon_1\\ \epsilon_2 \end{pmatrix}$ do not seem to me expressible as marginals of a common joint cdf

My idea is to define $$V\equiv \begin{pmatrix} \epsilon_1\\ \epsilon_2\\ \epsilon_1-\epsilon_2 \end{pmatrix}$$ with cdf $G$ and use $G$ as common cdf to re-write those probabilities. With the help of the picture above, $$\mathbb{P}(Y=1)=$$ $$\mathbb{P}( \epsilon_1\geq -u_1, \epsilon_2\leq -u_2)+ \mathbb{P}(\epsilon_1\geq -u_1, \epsilon_2\geq -u_2,\epsilon_1-\epsilon_2\geq u_2-u_1)$$ $$= \mathbb{P}( \epsilon_1\geq -u_1, \epsilon_2\leq -u_2,\epsilon_1-\epsilon_2\geq u_2-u_1)+ \mathbb{P}(\epsilon_1\geq -u_1, \epsilon_2\geq -u_2,\epsilon_1-\epsilon_2\geq u_2-u_1)$$ where the second equality comes from observing that $$\epsilon_1\geq -u_1, \epsilon_2\leq -u_2 \rightarrow \epsilon_1-\epsilon_2\geq u_2-u_1$$

Similarly, $$\mathbb{P}(Y=2)=$$ $$\mathbb{P}( \epsilon_1\leq -u_1, \epsilon_2\geq -u_2)+ \mathbb{P}(\epsilon_1\geq -u_1, \epsilon_2\geq -u_2,\epsilon_1-\epsilon_2\leq u_2-u_1)$$ $$= \mathbb{P}( \epsilon_1\leq -u_1, \epsilon_2\geq -u_2,\epsilon_1-\epsilon_2\leq u_2-u_1)+ \mathbb{P}(\epsilon_1\geq -u_1, \epsilon_2\geq -u_2,\epsilon_1-\epsilon_2\leq u_2-u_1)$$

where the second equality comes from observing that $$\epsilon_1\leq -u_1, \epsilon_2\geq -u_2 \rightarrow \epsilon_1-\epsilon_2\leq u_2-u_1$$

Lastly, $$\mathbb{P}(Y=0)=$$ $$\mathbb{P}( \epsilon_1\leq -u_1, \epsilon_2\leq -u_2, \epsilon_2\geq -u_2,\epsilon_1-\epsilon_2\leq u_2-u_1)+ \mathbb{P}( \epsilon_1\leq -u_1, \epsilon_2\leq -u_2, \epsilon_2\geq -u_2,\epsilon_1-\epsilon_2\geq u_2-u_1)$$ $$= \mathbb{P}( \epsilon_1\leq -u_1, \epsilon_2\leq -u_2, \epsilon_2\geq -u_2,\epsilon_1-\epsilon_2\leq \infty)$$ All these probabilities can be re-written using $G$. Is this correct?

• At first sight: no. If e.g. $X,Y$ have a common distribution then there is no way to express the probability of sets like $\{X<Y\}$ or $\{X+Y<c\}$ by means of an expression in $F_{X,Y}$. We can write them as integrals $\int[x<y]dF_{X,Y}(x,y)$ and $\int[x+y<c]dF_{X,Y}(x,y)$ where $[\text{condition on } x,y]$ takes value $1$ if the condition is satisfied, and takes value $0$ otherwise. There it stops. Btw, CDF's are important as fully determining the distribution, but often not well-behaved as tools for the expression of probabilities. So don't bother too much. – drhab Jul 30 '18 at 12:17
• @drhab Thank you. I have tried to elaborate a bit more my thoughts. If you have time could you tell me where/why my derivations are wrong? – TEX Jul 30 '18 at 12:53
• Only the blue one. – Did Jul 30 '18 at 13:51
• @Did thanks, why the pink and green are wrong? – TEX Jul 30 '18 at 14:01
• Because the CDF only deals with corners such as the blue one, and combinations of those, and clearly the pink and green areas are not of this type. – Did Jul 30 '18 at 14:03