Joint CDF Continuity Conditions Explained

For a joint cumulative distribution function (CDF) $F(x,y)$ of random variables $X$ and $Y$, a point $(x,y)$ is considered a point of continuity if the value of the CDF at that point equals the limit from below in both variables. Mathematically, this means:

$F(x, y) = \lim_{h \to 0^+} F(x-h, y) = \lim_{k \to 0^+} F(x, y-k) = \lim_{h \to 0^+, k \to 0^+} F(x-h, y-k)$

This condition is equivalent to ensuring there is no probability mass concentrated exactly at the point $(x,y)$ or on the lines leading up to it. The jump in the CDF at $(x,y)$, denoted $J(x,y)$, is given by:

$J(x,y) = F(x,y) - F(x^-, y^-)$

Where $F(x^-, y^-) = \lim_{h \to 0^+, k \to 0^+} F(x-h, y-k)$.

The jump can also be expressed using probabilities:

$J(x,y) = P(X = x, Y < y) + P(X < x, Y = y) + P(X = x, Y = y)$

For $(x,y)$ to be a point of continuity, the jump $J(x,y)$ must be zero. This requires the following three probabilities to be zero:

• $P(X = x, Y < y) = 0$
• $P(X < x, Y = y) = 0$
• $P(X = x, Y = y) = 0$

We now analyze the given options to see which ones guarantee these conditions.

Analyzing Sufficiency of Condition C

Condition C states: $P(X = x) = 0$ and $P(Y = y) = 0$.

Let's examine the implications:

• If $P(X = x) = 0$, this implies that the probability of $X$ taking the specific value $x$ is zero. This means $P(X = x, Y < y) = 0$, $P(X = x, Y = y) = 0$, and $P(X = x, Y > y) = 0$.
• Similarly, if $P(Y = y) = 0$, this implies $P(X < x, Y = y) = 0$, $P(X = x, Y = y) = 0$, and $P(X > x, Y = y) = 0$.

When both $P(X = x) = 0$ and $P(Y = y) = 0$ hold, we simultaneously satisfy:

• $P(X = x, Y < y) = 0$
• $P(X < x, Y = y) = 0$
• $P(X = x, Y = y) = 0$

These are precisely the conditions required for $J(x,y) = 0$, meaning $(x,y)$ is a point of continuity. Therefore, Condition C is sufficient.

Analyzing Sufficiency of Condition D

Condition D states: $P(X = x, Y \le y) = 0$ and $P(X \le x, Y = y) = 0$.

Let's break down these conditions:

• $P(X = x, Y \le y) = P(X = x, Y < y) + P(X = x, Y = y)$. If $P(X = x, Y \le y) = 0$, since probabilities are non-negative, it must be that $P(X = x, Y < y) = 0$ and $P(X = x, Y = y) = 0$.
• $P(X \le x, Y = y) = P(X < x, Y = y) + P(X = x, Y = y)$. If $P(X \le x, Y = y) = 0$, it must be that $P(X < x, Y = y) = 0$ and $P(X = x, Y = y) = 0$.

Combining the implications from both parts of Condition D, we get:

• $P(X = x, Y < y) = 0$
• $P(X < x, Y = y) = 0$
• $P(X = x, Y = y) = 0$

These are the direct requirements for the jump $J(x,y)$ to be zero. Thus, $(x,y)$ is a point of continuity. Therefore, Condition D is sufficient.