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Chapter 1: Problem 12

Let \(X\) have the pdf \(f(x)=3 x^{2}, 0

Short Answer

Expert verified
The expected value of \(X^{3}\) calculated in part(a), \(E(X^{3})\) will be \(\frac{3}{6} = 0.5\). The result of part(b) shows that \(Y=X^{3}\) is uniformly distributed in the interval (0,1). The expected value of \(Y\) calculated in part(c), \(E(Y)\) is also \(\frac{1}{2} = 0.5\). Therefore, the result obtained in part(c) is exactly the same as that obtained in part(a).

Step by step solution

01

Calculate E(X^3)

Since \(X\) is a continuous random variable with pdf \(f(x)=3x^{2}\), the expectation of \(X^{3}\) given by \(E(X^{3}) = \int_0^1 x^{3} f(x) dx = \int_0^1 x^{3} 3x^{2} dx = 3\int_0^1 x^{5} dx\). Calculate the integral to get the result. This can be done using the formula \[\int x^{n} dx = \frac{1}{n+1}x^{n+1} + C\] where \(n\) is a number, and \(C\) is the constant of integration.
02

Show Y = X^3 is Uniformly Distributed

Let \(Y = X^{3}\), the cumulative distribution function (cdf) of \(Y\) is given by \(F_Y(y) = P(Y \leq y) = P(X^{3} \leq y) = P(X \leq y^{1/3}) = F_X(y^{1/3})\). As \(X\) is distributed with pdf \(f(x)=3x^{2}\), this would imply \(F_X(x) = \int_{0}^{x} 3t^2 dt = x^3\). Substituting in the calculated value of \(F_X(x)\) to the equation of \(F_Y(y)\), the cdf of \(Y\) would become \(F_Y(y) = y\). All that's left is calculating the pdf of \(Y\) by taking the derivative of \(F_Y(y)\), where \(f_Y(y) = F_Y'(y)\), if \(f_Y(y) = 1\), this would mean that \(Y= X^{3}\) is uniformly distributed in the interval (0,1).
03

Compute E(Y) and Comparison with E(X^3)

For a random variable uniformly distributed in the interval (0,1), its expected value \(E(Y)\) is given by the formula \(E(Y) = \frac{1}{2}\). Compare this result with the result we obtained in Step 1 for \(E(X^{3})\).

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Key Concepts

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

Probability Density Function
The probability density function (pdf) is a fundamental concept in statistics that describes the likelihood of a continuous random variable taking on a specific value. In simpler terms, it defines how the values of the variable are distributed over possible values. The pdf can take on many forms, but the key requirement is that it must be non-negative everywhere, and the integral over the entire space must equal 1, indicating that the probability of the random variable falling within the range is certain.

For example, in our exercise the pdf of a random variable X is described by the function f(x) = 3x2 within the interval (0,1) and is zero elsewhere. This defines the behavior of X only within this range and tells us that the probability is greater for larger values of X closer to 1 than for values closer to 0.
Expected Value
The expected value, commonly denoted as E(X), is essentially the mean or the average that one would expect after observing a random variable many times over. Technically, it's the weighted average of all possible values that a random variable can take on, with the weights being their respective probabilities.

For continuous random variables, the expected value is calculated using integration. In our exercise, we computed the expected value of X3 by integrating the product of X3 and the pdf of X, over the range of X. This not only gives insight into the 'center of mass' of the distribution but is also a crucial part of understanding the behavior of the random variable.
Uniform Distribution
Uniform distribution is as straightforward as its name suggests; all outcomes are equally likely within a certain range. A uniform distribution can be described by two parameters, the lower limit a, and the upper limit b. In our case, we dealt with a uniform distribution over the interval (0,1), which means that every value between 0 and 1 is equally probable.

In the exercise, the transformation Y = X3 resulted in Y having a uniform distribution. This uniform distribution shows that the values between 0 and 1 are equally favored, and the graph of its pdf would be a flat line, indicating constant probability across its range.
Cumulative Distribution Function
Whereas pdf gives us the density of probabilities at any given point, the cumulative distribution function (cdf), denoted by F(x), tells us the probability that a random variable X is less than or equal to a particular value. Essentially, it's the accumulation of probabilities up to that point. For continuous distributions, the cdf is the integral of the pdf up to x, and it should be a non-decreasing, right-continuous function that approaches 0 as x approaches negative infinity and 1 as x approaches positive infinity.

Understanding cdf is crucial for handling probabilities over intervals, which can be visualized as the area under the pdf curve up to a certain point. In the context of the exercise, we used the cdf to demonstrate that Y takes on a uniform distribution by showing that the cdf of Y is a linear function that increases from 0 to 1 over the interval (0,1).

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Most popular questions from this chapter

Let the three mutually independent events \(C_{1}, C_{2}\), and \(C_{3}\) be such that \(P\left(C_{1}\right)=P\left(C_{2}\right)=P\left(C_{3}\right)=\frac{1}{4} .\) Find \(P\left[\left(C_{1}^{c} \cap C_{2}^{c}\right) \cup C_{3}\right]\)

A drawer contains eight different pairs of socks. If six socks are taken at random and without replacement, compute the probability that there is at least one matching pair among these six socks. Hint: Compute the probability that there is not a matching pair.

Players \(A\) and \(B\) play a sequence of independent games. Player \(A\) throws a die first and wins on a "six." If he fails, \(B\) throws and wins on a "five" or "six." If he fails, \(A\) throws and wins on a "four," "five," or "six." And so on. Find the probability of each player winning the sequence.

In an office there are two boxes of thumb drives: Box \(A_{1}\) contains seven 100 GB drives and three 500 GB drives, and box \(A_{2}\) contains two 100 GB drives and eight 500 GB drives. A person is handed a box at random with prior probabilities \(P\left(A_{1}\right)=\frac{2}{3}\) and \(P\left(A_{2}\right)=\frac{1}{3}\), possibly due to the boxes' respective locations. A drive is then selected at random and the event \(B\) occurs if it is a \(500 \mathrm{~GB}\) drive. Using an equally likely assumption for each drive in the selected box, compute \(P\left(A_{1} \mid B\right)\) and \(P\left(A_{2} \mid B\right)\)

Let \(X\) be a random variable with space \(\mathcal{D}\). For \(D \subset \mathcal{D}\), recall that the probability induced by \(X\) is \(P_{X}(D)=P[\\{c: X(c) \in D\\}] .\) Show that \(P_{X}(D)\) is a probability by showing the following: (a) \(P_{X}(\mathcal{D})=1\). (b) \(P_{X}(D) \geq 0\). (c) For a sequence of sets \(\left\\{D_{n}\right\\}\) in \(\mathcal{D}\), show that $$ \left\\{c: X(c) \in \cup_{n} D_{n}\right\\}=\cup_{n}\left\\{c: X(c) \in D_{n}\right\\} $$ (d) Use part (c) to show that if \(\left\\{D_{n}\right\\}\) is sequence of mutually exclusive events, then $$ P_{X}\left(\cup_{n=1}^{\infty} D_{n}\right)=\sum_{n=1}^{\infty} P_{X}\left(D_{n}\right) $$
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