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Chapter 2: Problem 53

If \(X\) is uniform over \((0,1)\), calculate \(E\left[X^{n}\right]\) and \(\operatorname{Var}\left(X^{n}\right)\).

Short Answer

Expert verified
The expected value of \(X^n\) is \(E(X^n) = \frac{1}{n+1}\), and the variance of \(X^n\) is \(\operatorname{Var}(X^n) = \frac{1}{2n+1} - \frac{1}{(n+1)^2}\).

Step by step solution

01

Calculate the Expected Value

To calculate the expected value \(E(X^n)\), we need to integrate the product of \(x^n\) and the probability density function \(f_X(x)\) over the support of the random variable \(X\) i.e., from \(0\) to \(1\). Using the definition of expectation for a continuous random variable, we have: \[E(X^n) = \int_0^1 x^n f_X(x) dx\] Since \(f_X(x) = 1\) for \(0 \le x \le 1\), the integral becomes: \[E(X^n) = \int_0^1 x^n dx\]
02

Solve the Integral

Now, let's solve the integral: \[\int_0^1 x^n dx = \frac{x^{n+1}}{n+1} \Big|_0^1 = \frac{1^{n+1}}{n+1} - \frac{0^{n+1}}{n+1}\] \[E(X^n) = \frac{1}{n+1}\] So, the expected value \(E(X^n) = \frac{1}{n+1}\).
03

Calculate the Second Moment

To calculate the variance, we also need the second moment of the random variable, that is \(E\left[(X^n)^2\right] = E(X^{2n})\). Following the same steps as before: \[E(X^{2n}) = \int_0^1 x^{2n} f_X(x) dx\] \[E(X^{2n}) = \int_0^1 x^{2n} dx = \frac{x^{2n+1}}{2n+1} \Big|_0^1 = \frac{1}{2n+1}\] So, the second moment of \(X^n\) is \(E(X^{2n}) = \frac{1}{2n+1}\).
04

Calculate the Variance

Using the formula for variance of a continuous random variable, we have: \[\operatorname{Var}(X^n) = E\left[(X^n)^2\right] - \left[E(X^n)\right]^2\] Plugging the values of \(E(X^n)\) and \(E(X^{2n})\): \[\operatorname{Var}(X^n) = \frac{1}{2n+1} - \left(\frac{1}{n+1}\right)^2\] So, the variance of \(X^n\) is \(\operatorname{Var}(X^n) = \frac{1}{2n+1} - \frac{1}{(n+1)^2}\).

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Key Concepts

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

Uniform Distribution
A uniform distribution is one where all outcomes in a given interval are equally likely. In the case of a continuous uniform distribution between 0 and 1, every value from 0 to 1 has the same chance of occurring. This is expressed with a constant probability density function.

For a random variable, say \(X\), that is uniformly distributed over \((0, 1)\), the probability density function \(f_X(x)\) is 1. This is because the total probability within the interval must sum to 1. Let鈥檚 break it down:
  • The total area under the probability density function curve from 0 to 1 is 1.
  • Every point within this range contributes equally to this area.
This concept is simple yet powerful, especially when calculating probabilities and expectations for continuous variables.
Expected Value
The expected value is a crucial concept in probability as it provides the average or mean of a random variable. It is the value you would "expect" to get, on average, if you repeated an experiment many times.

For a continuous random variable \(X\), the expected value \(E(X^n)\) can be calculated using integrals. Specifically, if \(X\) is uniformly distributed on \((0, 1)\), the formula becomes: \[E(X^n) = \int_0^1 x^n dx\] By solving this, we find: \[E(X^n) = \frac{1}{n+1}\]

This formula shows how the expected value changes with different powers \(n\) of the random variable \(X\). It鈥檚 a handy tool for predicting average outcomes over time.
Variance
Variance measures how much the values of a random variable differ from the expected value. It tells us the degree of spread in the values.

For calculating the variance of \(X^n\) where \(X\) is uniform on \((0,1)\), we first find the second moment, \(E(X^{2n})\):\[E(X^{2n}) = \int_0^1 x^{2n} dx = \frac{1}{2n+1}\]
This helps us determine the variance: \[\operatorname{Var}(X^n) = E(X^{2n}) - \left(E(X^n)\right)^2\]Substituting the known values:\[\operatorname{Var}(X^n) = \frac{1}{2n+1} - \frac{1}{(n+1)^2}\]
This calculation shows us how dispersed the values \(X^n\) are around the mean, and it varies depending on the power \(n\).
Continuous Random Variables
Continuous random variables can take any value within a given range, and are described by probability density functions. Unlike discrete random variables, which take on distinct values, continuous variables fill a continuum of possible outcomes.

In the case of a uniform distribution between 0 and 1, the variable can be any real number between these bounds. So:
  • Every possible number in range is technically an outcome.
  • Calculating probabilities involves finding the area under the curve of their probability density function.
Such variables are integral for modeling real-world phenomena where values are not distinct but rather flow within a range, like temperatures, heights, or time. Understanding their distributions, expected values, and variances is crucial for probabilistic predictions and analyses.

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Most popular questions from this chapter

An urn contains \(2 n\) balls, of which \(r\) are red. The balls are randomly removed in \(n\) successive pairs. Let \(X\) denote the number of pairs in which both balls are red. (a) Find \(E[X]\). (b) Find \(\operatorname{Var}(X)\).

If a fair coin is successively flipped, find the probability that a head first appears on the fifth trial.

Let \(X_{1}, X_{2}, \ldots, X_{10}\) be independent Poisson random variables with mean \(1 .\) (a) Use the Markov inequality to get a bound on \(P\left\\{X_{1}+\cdots+X_{10} \geq 15\right\\}\). (b) Use the central limit theorem to approximate \(P\left(X_{1}+\cdots+X_{10} \geq 15\right\\}\).

Suppose a die is rolled twice. What are the possible values that the following random variables can take on? (a) The maximum value to appear in the two rolls. (b) The minimum value to appear in the two rolls. (c) The sum of the two rolls. (d) The value of the first roll minus the value of the second roll.

A total of \(r\) keys are to be put, one at a time, in \(k\) boxes, with each key independently being put in box \(i\) with probability \(p_{i}, \sum_{i=1}^{k} p_{i}=1 .\) Each time a key is put in a nonempty box, we say that a collision occurs. Find the expected number of collisions.
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