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

Short Answer

Expert verified
The conditional expectation \(E\left[X \mid X<\frac{1}{2}\right]\) is \(\frac{1}{4}\).

Step by step solution

01

Identify the pdf of \(X\)

Since \(X\) is uniform over \((0,1)\), the pdf will be constant over this interval. The following is pdf of \(X\): \( p(x) = 1, \) for \(0 \leq x < 1 \), otherwise, \( p(x) = 0 \).
02

Determine the new range of \(X\) given the condition \(X < \frac{1}{2}\)

Given the condition \(X<\frac{1}{2}\), the new range of \(X\) will be from 0 to \(\frac{1}{2}\), inclusive of 0, but exclusive of \(\frac{1}{2}\). The pdf of \(X\) remains the same, but over this new range.
03

Find the conditional expectation

Now, we will calculate the conditional expectation \(E\left[X \mid X<\frac{1}{2}\right]\) which is the mean of the variable over the new range (0 to \(\frac{1}{2}\)). The formula for finding the mean in a uniform distribution over the range (a,b) is given by \(\frac{a+b}{2}\). Substituting in our values we get: \[E\left[X \mid X<\frac{1}{2}\right] = \frac{0+ \frac{1}{2}}{2} = \frac{1}{4}\]. So, the expected value of \( X \) given that \( X < \frac{1}{2} \) is \( \frac{1}{4} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Uniform Distribution
When you say that a variable like \( X \) is uniform over an interval, it means that the probability is evenly spread out across that range. In our case, with \( X \) uniform over (0,1), every value between 0 and 1 is equally likely.
This is known as uniform distribution, one of the simplest forms of probability distributions.
Key characteristics of a uniform distribution:
  • All outcomes within the interval are equally likely.
  • It's defined by only its minimum and maximum values.
  • The shape looks like a rectangle when graphed.
This simplicity makes uniform distribution a great starting point for learning probability concepts. It shows how probability can be equally allocated across a continuous range.
Probability Density Function
The probability density function (pdf) helps us understand how probabilities are spread across different values of a random variable.
For a continuous random variable like \( X \) with a uniform distribution over (0,1), the pdf is particularly simple: it's a constant value where the random variable could take a value within the defined interval.
In our case, since \( X \) is uniform over (0,1), its pdf \( p(x) \) is:
\[p(x) = 1, \text{ for } 0 \leq x < 1\]
When the variable \( X \) doesn't fall in the interval (e.g., less than 0 or equal to 1), \( p(x) \) becomes zero. The pdf is critical for calculating probabilities and expectations in continuous settings as it tells precisely how dense the probability is at each point across its range.
Expected Value
The expected value, or mean, of a probability distribution provides a measure of the "central" or average magnitude of the random variable. It encapsulates the long-term average or mean value if you could observe the outcome of \( X \) many times.
For a uniformly distributed \( X \) over (0,1), the expected value across its full range can be calculated simply as the midpoint, \( \frac{1}{2} \).
However, when conditions like \( X < \frac{1}{2} \) are introduced, we calculate the conditional expected value, which considers only this subset of \( X \)'s range. Here, we have defined the new range as (0, \( \frac{1}{2} \)) and used the formula for the mean of \( X \), shrinking the perspective accordingly:
\[ E\left[X \mid X<\frac{1}{2}\right] = \frac{0 + \frac{1}{2}}{2} = \frac{1}{4} \]
This approach reflects how the average value shifts when we look at a specific part of a distribution.
Conditional Probability
Conditional probability is the probability of an event occurring given that another event has already occurred. In the context of expected value and probability distributions, it helps us zoom in to assess how statistics change under given conditions.
Here, we deal with the condition \( X < \frac{1}{2} \). This means we are interested in only the values of \( X \) that are less than \( \frac{1}{2} \).
This transforms the uniform distribution to focus solely on the interval (0, \( \frac{1}{2} \)). The probability densities and expected value (mean) calculation become restricted to this slice of the original distribution. Hence, conditional expectation \( E[X \mid X<\frac{1}{2}] \) represents the average of \( X \) within this condition, capturing how expectations adjust when considering specific events or ranges.

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Most popular questions from this chapter

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