I read in my statistics book about Discrete Uniform Probability Distribution and the Expected Value
A discrete random variable $X$ with $k$ possible outcomes $x_1, x_2, …,x_k$ is said to follow a discrete uniform distribution if the probability mass function
(PMF) of $X$ is given by
$$P(X = x_i) = \frac{1}{k}, \forall i = 1,2,…,k $$
If the outcomes are the natural numbers $x_i = i (i = 1, 2, . . . , k)$, the mean and variance of $X$ are obtained as
$$E(X) = \frac{k+1}{2}, $$
$$Var(X) = \frac{1}{12}(k^2 – 1) $$
I have a question on $E(X)$. It seems I find it hard to understand the formula for $E(X)$ for coin toss example. in a fair coin toss we have $k = 2$ (heads or tails, here $0$ and $1$ respectivley) and $ E(X)= \frac{2+1}{2} = 1.5$
However, when you calculate the $E(X)$ as
$ E(X) = $ probability weighted sum of $k$ it is
$$ E(X) = 0\times0.5 + 1\times0.5 = 0.5 $$
So why is this difference? What am I missing?
Best Answer
The way you define the random variable $X$ is not consistent between the two approaches, which is why you have different results. In the discrete uniform case, the support is $X \in \{1, 2\}$, with probability mass function $$\Pr[X = 1] = 1/2, \\ \Pr[X = 2] = 1/2.$$ Hence the expectation is $$\operatorname{E}[X] = 1 \cdot \Pr[X = 1] + 2 \Pr[X = 2] = \frac{1}{2} + \frac{2}{2} = \frac{3}{2}.$$ When you try to represent the outcomes of a fair coin toss with $X$, you instead wrote $X \in \{0, 1\}$, hence $$\Pr[X = 0] = 1/2, \\ \Pr[X = 1] = 1/2.$$ Obviously, you will not get the same result.