The expected value (or expectation) of a discrete random variable is essentially the "long-run average" value of that variable.
Given a discrete random variable $X$ with a probability mass function $P(X)$, the expectation of a function $g(X)$ is defined as:
$$\mathbb{E}[g(X)] = \sum_{x \in \text{Val}(X)} g(x)P(x)$$
Where:
When the function $g(X)$ is simply the identity $g(X) = X$, the expectation gives us the mean ($\mu$) of the distribution:$$\mu = \mathbb{E}[X] = \sum_{x} x \cdot P(x)$$
Imagine a robot is moving in a 1D hallway with 5 discrete cells. Its current probability distribution (the "belief") for its position $X$ is as follows:
| Position ($x$) | $P(x)$ | $x \cdot P(x)$ |
|---|---|---|
| 1 | 0.05 | 0.05 |
| 2 | 0.10 | 0.20 |
| 3 | 0.70 | 2.10 |
| 4 | 0.10 | 0.40 |
| 5 | 0.05 | 0.25 |
| Total | 1.00 | $\mathbb{E}[X] = 3.0$ |
To find the expected position of the robot, we calculate the weighted average:
$$\mathbb{E}[X] = (1 \cdot 0.05) + (2 \cdot 0.10) + (3 \cdot 0.70) + (4 \cdot 0.10) + (5 \cdot 0.05)$$
$$\mathbb{E}[X] = 0.05 + 0.20 + 2.10 + 0.40 + 0.25 = \mathbf{3.0}$$
Interpretation: Even though the robot has a small chance of being at any of the 5 positions, its "expected" location is exactly at cell 3. This value acts as the "center of mass" of the probability distribution.
For a continuous random variable $X$, we replace the discrete summation with an integral over the entire domain of $X$. If $p(x)$ is the probability density function (PDF), the expectation of a function $f(X)$ is defined as:
Expectation of a continuous random variable
$$\mathbb{E}[f(X)] = \int_{-\infty}^{\infty} f(x) p(x) \, dx$$
Just as in the discrete case, the mean of a continuous distribution is the expected value of the identity function $f(x) = x$:$$\mu = \mathbb{E}[X] = \int_{-\infty}^{\infty} x p(x) \, dx$$
Imagine a robot is located somewhere in a 10-meter-long tunnel, and its position is represented by a Uniform Distribution $U(0, 10)$. The probability density function is:
$$p(x) = \begin{cases} \frac{1}{10} & \text{if } 0 \leq x \leq 10 \\ 0 & \text{otherwise} \end{cases}$$
To find the expected position:
$$\mathbb{E}[X] = \int_{0}^{10} x \cdot \frac{1}{10} \, dx$$
Solving the integral:
$$\mathbb{E}[X] = \frac{1}{10} \left[ \frac{1}{2}x^2 \right]_0^{10} = \frac{1}{10} \left( \frac{100}{2} - 0 \right) = \mathbf{5.0}$$
Interpretation: The expected value is the geometric center of the distribution. In robotics, if you have no information other than the boundaries of the environment, the expectation is the "least wrong" guess.
| Feature | Discrete | Continuous |
|---|---|---|
| Operator | Summation ($\sum$) | Integration ($\int$) |
| Probability | Mass Function (PMF) | Density Function (PDF) |
| Definition | $\mathbb{E}[f(x)] = \sum f(x)p(x)$ | $\mathbb{E}[f(x)] = \int f(x)p(x) dx$ |
| Total Probability | $\sum p(x) = 1$ | $\int p(x) dx = 1$ |