Question

A continuous random variable \(X\) has PDF \(f(x)=\) \(\frac{3}{256} x(8-x), 0 \leq x \leq 8 .\) Find \(E\left(X^{2}\right)\) and \(E\left(X^{3}\right)\).

Short answer

\(E(X^2) \approx 19.21\) and \(E(X^3) \approx -204.80\).

Step-by-step solution

Understand the Problem

We are given a probability density function (PDF) for a continuous random variable \(X\). The PDF is \(f(x) = \frac{3}{256} x(8-x)\) for \(0 \leq x \leq 8\). Our task is to find the expected values \(E(X^2)\) and \(E(X^3)\). The expected value for any function of \(X\), say \(g(X)\), is given by \(E(g(X)) = \int_{-fty}^{fty} g(x) f(x) \, dx\). Since the function is only defined between \(0\) and \(8\), our integration limits are from \(0\) to \(8\).

Set Up the Integral for E(X^2)

The expected value \(E(X^2)\) is calculated by integrating \(x^2 f(x)\) over the range from \(0\) to \(8\). This gives:\[E(X^2) = \int_{0}^{8} x^2 \frac{3}{256} x(8-x) \, dx = \int_{0}^{8} \frac{3}{256} x^3(8-x) \, dx.\]

Simplify the Integral for E(X^2)

Expand the expression in the integrand:\[\int_{0}^{8} \frac{3}{256} (8x^3 - x^4) \, dx.\]This becomes two separate integrals:\[\frac{3}{256} \left( \int_{0}^{8} 8x^3 \, dx - \int_{0}^{8} x^4 \, dx \right).\]

Calculate Separate Integrals for E(X^2)

Calculate each integral separately:- \(\int_{0}^{8} 8x^3 \, dx = \left[ 2x^4 \right]_{0}^{8} = 2(8^4) - 0 = 2 \times 4096 = 8192.\)- \(\int_{0}^{8} x^4 \, dx = \left[ \frac{x^5}{5} \right]_{0}^{8} = \frac{8^5}{5} - 0 = \frac{32768}{5} = 6553.6.\)

Combine Results to Find E(X^2)

Substitute the results back into the integral formula:\[E(X^2) = \frac{3}{256} (8192 - 6553.6) = \frac{3}{256} \times 1638.4 = 19.21425.\]

Set Up the Integral for E(X^3)

Similarly, calculate \(E(X^3)\) by integrating \(x^3 f(x)\) over the range from \(0\) to \(8\):\[E(X^3) = \int_{0}^{8} x^3 \frac{3}{256} x(8-x) \, dx = \int_{0}^{8} \frac{3}{256} x^4(8-x) \, dx.\]

Simplify the Integral for E(X^3)

Expand the expression in integrand:\[\int_{0}^{8} \frac{3}{256} (8x^4 - x^5) \, dx.\]This becomes two separate integrals:\[\frac{3}{256} \left( \int_{0}^{8} 8x^4 \, dx - \int_{0}^{8} x^5 \, dx \right).\]

Calculate Separate Integrals for E(X^3)

Calculate each integral separately:- \(\int_{0}^{8} 8x^4 \, dx = \left[ \frac{8x^5}{5} \right]_{0}^{8} = \frac{8 \times 32768}{5} = 52428.8.\)- \(\int_{0}^{8} x^5 \, dx = \left[ \frac{x^6}{6} \right]_{0}^{8} = \frac{8^6}{6} = 69904.667.\)

Combine Results to Find E(X^3)

Substitute the results back into the integral formula:\[E(X^3) = \frac{3}{256} (52428.8 - 69904.667) = \frac{3}{256} \times (-17475.867) = -204.7995.\]

Conclusion: Final Results

The expected value \(E(X^2)\) is approximately 19.21 and \(E(X^3)\) is approximately -204.80.

Key concepts

Continuous Random Variable

A continuous random variable lies at the core of probability and statistics. It is defined as a variable that can take an infinite number of possible values in a given range. Unlike discrete variables, which have specific set outcomes, continuous random variables cover any value between two numbers. For instance, if you're measuring height, the range could be between say 150 cm and 200 cm. And within this range, the height can be virtually any value, like 172.3 cm or even 172.322 cm.
The probability distribution of a continuous random variable is typically described by a probability density function (PDF). The PDF gives us how probabilities are distributed over values of the random variable. The area under the curve of the PDF over an interval gives the probability that the variable falls within that interval. For example, the PDF for our variable is defined as \(f(x) = \frac{3}{256} x(8-x)\) over the interval \(0 \leq x \leq 8\). This means that the probability characteristics reset outside the 0 to 8 range.

Expected Value

Expected value is a fundamental concept in probability and statistics used to determine the average outcome of a random event. For a continuous random variable \(X\), the expected value \(E(X)\) is essentially the long-run average value after many trials of the random process.
To calculate the expected value of a function of \(X\), you integrate this function multiplied by the probability density function over the possible values of \(X\). This is formulated as: \[E(g(X)) = \int_{-\infty}^{\infty} g(x) f(x) \, dx\] However, for the defined range \(0 \leq x \leq 8\), this simplifies to: \[E(g(X)) = \int_{0}^{8} g(x) f(x) \, dx\] This way, you can find expected values for powers of \(X\), such as \(E(X^2)\) and \(E(X^3)\), by setting \(g(x)\) respectively to \(x^2\) and \(x^3\) and integrating over the function's range.

Integral Calculus

Integral calculus is a branch of mathematics that deals with integrals and their properties. It encompasses two main mathematical operations: determining the area under a curve (definite integrals) and finding antiderivatives (indefinite integrals).
For probability and continuous random variables, integral calculus is crucial. The expected value calculations for continuous random variables are performed using definite integrals, which involve finding the area under the curve described by the function of \(X\) multiplied by its PDF.
For example, when calculating \(E(X^2)\) for a given \(X\) with PDF \(f(x)\), you compute: \[E(X^2) = \int_{0}^{8} x^2 f(x) \, dx.\] This involves expanding the integrand, calculating separate integrals, and combining those results. Without integral calculus, defining exact probabilities and expected values for continuous distributions would be nearly impossible. This mathematical tool enables a systematic approach to aggregating infinite values that a continuous random variable can assume.