# Discrete random variables and independent events

Let $X$ be a discrete random variable. The probability function is given in the following table

$$\begin{array}{c|cccc} x&-10&1&5&10\\ \hline f(x)&0.1&0.3&0.4&0.2 \end{array}$$

I wish to compute probability function for discrete random variable $Y$ such that $Y=X^2$

If I want to know $f(100)$, then

\begin{align*} f(100)=P(Y=100)=P(X^2=100)&=P(\{X=10\}\cup\{X=-10\})\\ &=P(X=10)+P(X=-10)-P(\{X=10\}\cap\{X=-10\})\\ &=0.2+0.1-0\\ &=0.3 \end{align*} Why the events $\{X=10\}$ and $\{X=-10\}$ are independent?

They are not independent, they are mutually exclusive. Events $A$ and $B$ are independent if $$P(A \cap B) = P(A) P(B)$$ $A$ and $B$ are mutually exclusive if $$P(A \cap B) = 0$$ $\{X = 10\}$ and $\{X = -10\}$ are mutually exclusive since we can't simultaneously have $X = 10$ and $X = -10$.
• Why can not we have simultaneously $X=10$ and $X=-10$? – user548718 Apr 4 '18 at 12:44
• Because $\{X = 10\}$ and $\{X = -10\}$ are different events. $X$ can only have one value. It's like you can't have a coin flip be simultaneously heads and tails. An example of two events that aren't mutually exclusive are $\{X = 10\ \text{or}\ X = 1\}$ and $\{X = 1\ \text{or}\ X = -10\}$, since if $X = 1$ both events occur. – bitesizebo Apr 4 '18 at 12:46
No, they aren't independent because $$P(\{X=10\} \cap \{X=-10\}) \ne P(\{X=10\}) P(\{X=-10\}).$$
• $P(\{X=10\} \cap \{X=-10\}) = P(\varnothing) = 0$
• $P(\{X=10\}) P(\{X=-10\}) = 0.2 \times 0.1 > 0$
• Why $P(\{X=10\} \cap \{X=-10\}) = P(\varnothing) = 0$? – user548718 Apr 4 '18 at 12:45
• @ThomasHiler $$\{X=10\} \cap \{X=-10\} = \varnothing$$ since $X: \Omega \to \Bbb{R}$ is a function, so it can't map $\omega \in \Omega$ into two disjoint subset of $\Bbb{R}$. – GNUSupporter 8964民主女神 地下教會 Apr 4 '18 at 12:47