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The expected value is a fundamental concept in probability, often considered the "average" or "mean" of a random variable. It provides a measure of the center of the distribution of a random variable.
The expected value of a continuous random variable, like the one given in our problem, involves integrating over a probability density function (pdf). This is different from the discrete version, where we sum over possible outcomes.

In mathematical terms, for a continuous random variable with a pdf of \( f(x) \), the expected value \( E(X) \) is calculated by:

• Computing the integral \( E(X) = \int_{a}^{b} x \, f(x) \, dx \).

The concept is similar for finding the expected value of a transformation of \( X \), which in our case is \( \frac{1}{X} \).

Here's what we do:

• Apply the formula \( E(g(X)) = \int_{a}^{b} g(x) \, f(x) \, dx \).
• For \( g(x) = \frac{1}{x} \) with a uniform pdf over \([1, 2] \), we solve \( E\left( \frac{1}{X} \right) = \int_{1}^{2} \frac{1}{x} \, dx \).
• This integral evaluates to \( \ln 2 \), showing us that the expected value of \( \frac{1}{X} \) is \( \ln 2 \).

The expected value allows us to understand the long-term behavior of transformations of random variables.

The concept of a probability density function (pdf) is crucial when dealing with continuous random variables. It describes the likelihood of a random variable to take on a particular value.
For continuous random variables, unlike discrete ones, probabilities of exact values are zero. Instead, probabilities are assigned over intervals.

In the case of a uniform distribution on an interval \([a, b]\):

• The pdf is given by \( f(x) = \frac{1}{b-a} \) for \( a \leq x \leq b \).
• This means the probability is evenly spread across the interval.

In our problem, the pdf for \( X \) is \( f(x) = 1 \) for \( x \) in \([1, 2]\), showing equal likelihood across the interval.

This simplification arises because:

• The total area under the pdf, representing total probability, must be 1.
• A uniform distribution has a constant pdf across its interval.

The role of the pdf in calculus comes into play through integrals, allowing us to find expected values and other properties of the distribution.

A continuous random variable is one that can take on any value within a given range. This characteristic differentiates it from discrete random variables, which take on distinct or separate values.
In many cases, continuous data can provide a more accurate representation of natural phenomena.

Key properties include:

• Can assume an infinite number of potential values within a specified range.
• Typically described using probability density functions (pdf), not probability mass functions (pmf).
• Probabilities of exact values are zero, hence we consider probabilities over intervals.

For a uniformly distributed continuous variable, like \( X \) in our problem:

• It is uniformly spread over \([1, 2]\).
• The probability of \( X \) taking any specific value within this range is equally likely.

This property of uniform distribution is pivotal when calculating probabilities and expected values, as it ensures a constant transition likelihood throughout the interval.