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A random variable is a numerical outcome of a random phenomenon. Think of it as a way to assign a number to every possible outcome of an event.
There are different types of random variables, such as discrete and continuous.

• Discrete Random Variables: These take on a countable number of distinct values, like rolling a die or flipping a coin.
• Continuous Random Variables: These can take on any value within a certain range, like the height of a person or the amount of rainfall in a day.

Each random variable is associated with a probability distribution, which tells us the likelihood of different outcomes.

The expectation (or mean) of a random variable is a measure of its average value over many trials.
When dealing with a constant random variable, where every outcome is the same, the concept becomes straightforward.
For a constant random variable defined as \(X(\omega) = a\) for all \(\omega\), the expectation is just the constant value \(a\).
This is because the expected value is essentially a weighted average, and if all weights are the same (i.e., the same constant value), then the average is simply that value.

• If \(X(\omega) = 5\), then \(E[X] = 5\).
• Similarly, if \(X(\omega) = 42\), then \(E[X] = 42\).

Consider a random variable representing the distance to the nearest endpoint of an interval.
For example, let \(X(\omega) = \min \{\omega, 1-\omega\}\) over the interval \([0,1]\). In this case, the expression \(\min \{\omega, 1-\omega\}\) tells us the minimum distance to either endpoint 0 or 1.
To find the expectation for this, we note that the function \(X(\omega)\) is symmetric around 0.5.
This symmetry simplifies the integration process because we only need to integrate from 0 to 0.5 and then double the result. To compute this:

• Integrate \(x\) from 0 to 0.5: \[\int_0^{0.5} x \, dx = \left[ \frac{x^2}{2} \right]_0^{0.5} = \frac{0.5^2}{2} = \frac{0.25}{2} = 0.125\]
• Double the result: \(2 \times 0.125 = 0.25\)

Thus, the expectation \(E[X] = 0.25\).

Symmetry in integrals can make complex problems easier to solve.
When a function is symmetric, we can often reduce the problem's scope and then apply the symmetry to find the solution. Consider the function \(X: [0,1]^2 \rightarrow \mathbb{R}\) representing the distance to the nearest edge of a square. If the point is \((x,y)\) in the square, the distance to the nearest edge is \(\min \, \{x, 1-x, y, 1-y\}\).
Taking advantage of symmetry in the interval [0,0.5] for both \(x\) and \(y\), we simplify the integral:

• Integrate \(x\) over [0,0.5]: \[4 \int_0^{0.5} \int_0^{0.5} x \, dy \, dx = 4 \int_0^{0.5} \left[ x y \right]_0^{0.5} dx = 4 \int_0^{0.5} x \times 0.5 \, dx\]
• Simplify and integrate: \[4 \times 0.5 \int_0^{0.5} x \, dx = 4 \times 0.5 \times \frac{x^2}{2} \Bigg|_0^{0.5} = 4 \times 0.5 \times 0.125 = 0.25\]

The expectation \(E[X] = 0.25\).
This example shows how symmetry can significantly simplify the calculation of integrals, especially in probability and expectation calculations.