Question

Let \(X\) be uniform over \((0,1) .\) Find \(E\left[X \mid X<\frac{1}{2}\right]\).

Short answer

The conditional expectation of \(X\) given that \(X < \frac{1}{2}\) is \(E\left[X \mid X<\frac{1}{2}\right] = \dfrac{1}{4}\).

Step-by-step solution

Find the conditional pdf of X given X < 1/2

Since \(X\) is uniformly distributed over \((0, 1)\), the pdf of \(X\), denoted as \(f_X(x)\), is: \(f_X(x) = \left\{ \begin{array}{ll} 1 & \textrm{for}\ 0 < x < 1 \\ 0 & \textrm{otherwise.} \end{array} \right.\) Now, we need to find the conditional pdf of \(X\) given that \(X < \frac{1}{2}\). For this, we need to use the definition of conditional probability: \(f_{X|X<\frac{1}{2}}(x) = \dfrac{f_X(x) \cdot \bold{1}(x < \frac{1}{2})}{P(X<\frac{1}{2})}\) where \(\bold{1}(A)\) denotes the indicator function that takes the value 1 if event \(A\) happens and 0 otherwise. Since \(X\) has a uniform pdf, we have \(P(X < \frac{1}{2}) = \int_0^{1/2} f_X(x) dx = \frac{1}{2}\) Applying this to find the conditional pdf: \(f_{X|X<\frac{1}{2}}(x) = \dfrac{f_X(x) \cdot \bold{1}(x < \frac{1}{2})}{\frac{1}{2}}\) This simplifies to: \(f_{X|X<\frac{1}{2}}(x) = \left\{ \begin{array}{ll} 2 & \textrm{for}\ 0 < x < \frac{1}{2} \\ 0 & \textrm{otherwise.} \end{array} \right.\)

Find E[X | X < 1/2] using the conditional pdf

Now that we have found the conditional pdf of \(X\) given that \(X < \frac{1}{2}\), we can use it to find the conditional expectation. Recall that for any random variable \(Y\) with pdf \(f_Y(y)\), the expectation is defined as: \(E[Y] = \int_{-\infty}^{\infty} y \cdot f_Y(y) dy\) In our case, we want to find \(E\left[X \mid X<\frac{1}{2}\right]\), using the conditional pdf \(f_{X|X<\frac{1}{2}}(x)\): \(E\left[X \mid X<\frac{1}{2}\right] = \int_{-\infty}^{\infty} x \cdot f_{X|X<\frac{1}{2}}(x) dx\) Using the expression for the conditional pdf \(f_{X|X<\frac{1}{2}}(x)\), we can simplify the integral as follows: \(E\left[X \mid X<\frac{1}{2}\right] = \int_{0}^{\frac{1}{2}} x \cdot 2 dx\) Now, we can integrate and find the conditional expectation: \(E\left[X \mid X<\frac{1}{2}\right] = 2\int_{0}^{\frac{1}{2}} x dx = 2\left[\dfrac{x^2}{2}\right]_0^{\frac{1}{2}} = \dfrac{(\frac{1}{2})^2 - 0^2}{1} = \dfrac{1}{4}\) So the conditional expectation of X given that X < 1/2 is \(E\left[X \mid X<\frac{1}{2}\right] = \dfrac{1}{4}\).

Key concepts

Uniform Distribution

The uniform distribution is often considered one of the simplest probability distributions in statistics. It is used to model scenarios where all outcomes are equally likely. If we say a random variable \(X\) has a uniform distribution over an interval \((a, b)\), this means that the probability is evenly spread across the interval, and every value between \(a\) and \(b\) is equally likely to occur.

To visualize, imagine a straight horizontal line on a graph from \(a\) to \(b\); this line represents the probability density function (pdf) for a uniform distribution. In the context of the provided exercise, \(X\) is uniform over \((0,1)\), and therefore the pdf of \(X\), \(f_X(x)\), is constant (equal to 1) between 0 and 1 and 0 elsewhere. This uniformity simplifies calculations involving probabilities and expectations, such as finding the expectation of \(X\) subjected to a condition like \(X<\frac{1}{2}\).

Probability Density Function

The probability density function (pdf) is at the heart of continuous probability distributions. It describes the likelihood of a random variable to take on a given value. Essentially, the pdf gives us a function where the area under the curve between two points corresponds to the probability that the random variable falls between those two values.

In formal terms, for a continuous random variable \(X\), the pdf \(f_X(x)\) is such that for any two numbers \(a\) and \(b\) where \(a < b\), the probability that \(X\) falls between \(a\) and \(b\) is given by the integral of \(f_X(x)\) from \(a\) to \(b\). It's critical to remember that for a pdf, the total area under the curve should be 1, symbolizing the certainty that \(X\) will take some value within its range. For the uniform distribution, this concept is straightforward since the pdf is a constant value where the random variable is defined.

Indicator Function

The indicator function is a simple yet powerful tool in probability and statistics. Denoted commonly by the symbol \(\bold{1}(A)\), where \(A\) is some event, it 'indicates' whether the event \(A\) has occurred or not. The function takes on a value of 1 if the event \(A\) happens and 0 otherwise.

For instance, in the problem at hand, \(\bold{1}(x < \frac{1}{2})\) is the indicator function that is 1 if \(x\) is less than \(\frac{1}{2}\) and 0 if it's not. It effectively 'turns on' the probability density function \(f_X(x)\) in the interval we're interested in and 'turns it off' outside of it. Indicator functions are very useful when dealing with conditional probabilities because they help to incorporate the condition directly into the probability expressions, allowing for a straightforward integration to calculate probabilities or expectations.