Question

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

Short answer

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

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.

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).

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 \(E(X^{3})\).

Key concepts

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).