CDF of $X+Y$ does not equal CDF of $X$ plus CDF of Y

Question

I am studying probability theory and I am very new with the concept.

In a tutorial exercise I need to show that the following statement is false:

If $X$ and $Y$ are two random variables defined on a probability space $(Ω, A , P)$, then $F_{X +Y} = F_X + F_Y $.

I tried to make a counterexample, but I am not sure if I compute the CDF of $X+Y$ correctly. Can someone verify this for me?

I came up with the following:

Consider two fair dices: one dice is just a normal dice (1,2,3,4,5,6), the other one has repetitive numbers (1,1,2,2,3,3).

Now I want to show that $F_{X+Y}(3) = F_X(3) + F_Y(3)$ doesn't hold:

Note: $F_X(3) = \frac{1}{2}$ and $F_Y(3) = 1$.

Now compute $F_{X+Y}(3)$:

$F_{X+Y}(3) = \frac{1}{6}*0+\frac{1}{6}*\frac{1}{3}+\frac{1}{6}*\frac{2}{3}+\frac{1}{6}*1+\frac{1}{6}*1+\frac{1}{6}*1 = \frac{2}{3}$

Thus, since $\frac{1}{2} +1 \neq \frac{2}{3}$, we have a contradiction here.

Is the way I computed $F_{X+Y}(3)$ correct?

Thanks in advance! :)

Answer 1

Let $Z = X+Y$. Now observe the following two quantities:

$$P(Z \le z) (1)$$

$$P(X \le z) + P(Y \le z) \quad (2)$$

Does $(2)$ imply $(1)$? No.

You can construct multiple counter examples without explicitly constructing a CDF. For example $X = z-1$ and $Y=z-1$. In general, the CDF for $Z$ will be calculated as follows in case $X$ and $Y$ are discrete RV:

$$F_Z(z) = P(Z \le z)$$ $$= \sum_{y}P(Z\le z|Y=y)\times P(Y=y)$$ $$=\sum_{y}P(X\le z-y)\times P(Y=y)$$ $$=\sum_{y}F_X(z-y) \times P(Y=y)$$

In case of continuous random variables, the following argument applies:

$$F_Z(z) = P(Z \le z)$$ $$= \int_{y} P(Z \le z|Y=y) \times f_Y(y) dy$$ $$=\int_{y} P(X \le z-y) \times f_Y(y) dy$$ $$=\int_{y} F_{X}(z-y) \times f_Y(y) dy$$

Answer 2

There isn't a need to compute $F_{X+Y}(3)$ actually since the sum of $F_X(3)$ and $F_Y(3)$ already exceeds $1$.

\begin{align} F_{X+Y}(3) &= Pr(X+Y=3) + Pr(X+Y=2) \\ &= Pr(X=2,Y=1)+Pr(X=1,Y=2)+ Pr(X=1,Y=1)\\ &= \frac16\cdot \frac13 + \frac16 \cdot \frac13 + \frac16\cdot \frac13 \\ &= \frac16 \end{align}

Answer 3

Sum of two distribution functions is never a distribution function. This is because any distribution function tends to 1at $\infty$ and $1+1\neq 1$