Question

Let \(X\) have the \(\operatorname{cdf} F(x)\) that is a mixture of the continuous and discrete types, namely $$ F(x)=\left\\{\begin{array}{ll} 0 & x<0 \\ \frac{x+1}{4} & 0 \leq x<1 \\ 1 & 1 \leq x \end{array}\right. $$ Determine reasonable definitions of \(\mu=E(X)\) and \(\sigma^{2}=\operatorname{var}(X)\) and compute each. Hint: Determine the parts of the pmf and the pdf associated with each of the discrete and continuous parts, and then sum for the discrete part and integrate for the continuous part.

Short answer

The expectation and variance for the mixed cummulative distribution function \(F(x)\) are \(\mu=E(X) = \frac{1}{8}\) and \(\sigma^2 = var(X) = \frac{1}{96}\) respectively.

Step-by-step solution

Determine the parts of the PDF

Looking at the \(F(x)\), the parts of the probability density function (pdf) can be determined. On differentiating \(F(x)\) with respect to \(x\), for \(0 \leq x < 1\), \(\frac{df}{dx}= \frac{1}{4}\). This gives the pdf \(f(x)\) for a continuous part of the distribution. For \(x >= 1\), \(F(x)\) is a constant, so \(\frac{df}{dx} = 0\), giving the pdf \(f(x)=0\) for the discrete part of the distribution.

Calculate the expectation for individual parts

The expectation \(E(X)\) can be calculated by multiplying each \(x\) value by its pdf. For \(0 \leq x < 1\), \(E(X) = \int_0^1 x \cdot \frac{1}{4} dx = \frac{1}{8}\). For \(x\ge1\), \(E(X) = \int_1^\infty x \cdot 0 dx = 0\). The total expectation is the sum of the two parts, \(\mu=E(X) = \frac{1}{8} + 0 = \frac{1}{8}\).

Compute the variance for individual parts

The variance \(var(X)\) can be calculated as the expectation of the squares minus the square of the expectation. For \(0\le x<1\), \(\mu_2 = E(X^2) =\int_0^1 x^2 \cdot \frac{1}{4} dx = \frac{1}{12}\), and the variance \(var(X) = \mu_2 - \mu^2 = \frac{1}{12} - \left(\frac{1}{8}\right)^2 = \frac{1}{96}\). For \(x \ge 1\), since \(f(x) = 0\), \(E(X^2) = \int_1^\infty x^2 \cdot 0 dx = 0\), and the variance \(var(X) = 0 - 0 = 0\). The total variance is the sum of the two parts, \(\sigma^2 = var(X) = \frac{1}{96} + 0 = \frac{1}{96}\).

Key concepts

Probability Density Function

When understanding how to analyze random phenomena, one of the key concepts is the probability density function (PDF). This function is vital for continuous random variables, as it describes the likelihood of a variable taking on a particular value. Unlike a cumulative distribution function (CDF), which tells us the probability that a variable will take a value less than or equal to a certain amount, the PDF tells you the 'density' of probability at any given point.

For instance, in the step-by-step solution provided, differentiation of the CDF yielded the PDF of the variable for the continuous part. The function was defined as \(f(x) = \frac{1}{4}\) for the interval \(0 \leq x < 1\), which means within this range, the variable has a uniform probability distribution. Simply put, across this interval, every value has an equal chance of occurring. This aspect is critical for calculating other statistics, such as the expectation and variance of the random variable.

In practice, if you wanted to calculate the probability of the random variable falling within a specific interval within the continuous range, you would integrate the PDF over that interval. However, keep in mind that for the discrete part, with \(x \geq 1\), the probability mass function (PMF) would instead be applied. With the given CDF, the PDF is determined to be 0 for \(x \geq 1\), indicating there's no 'density' of probability past the point \(x=1\).

Expectation of Random Variable

The expectation of a random variable, commonly denoted as \(E(X)\) or \(\mu\), is essentially a measure of the 'average' outcome you would expect if you were able to repeat an experiment an infinite number of times. In other words, it provides the central or 'expected' value around which the values of a random variable are distributed. For continuous distributions, the expectation is computed by integrating the product of the variable and its probability density function.

In the given solution, calculating the expectation involved integrating \(x\) multiplied by its probability density, over the interval from 0 to 1. The result, \(\mu = \frac{1}{8}\), represents the average value of the random variable over this continuous part. It's important to notice that the expectation for the discrete part after \(x \geq 1\) is zero because the PDF is zero there. This value implicates that the random variable’s most 'expected' or average outcomes lie within the continuous part of the distribution.

Variance of Random Variable

Studying the variance of a random variable, denoted as \(\sigma^2\) or \(\text{var}(X)\), enables one to understand the dispersion, or how much the values of this variable spread out from the expected value. A higher variance means that the values of the variable are more scattered, while a lower variance indicates they are closer to the mean (\(\mu\)).

To calculate variance for a continuous random variable, the expected value of the squared variable \(E(X^2)\) is found by integrating the square of the variable times the PDF over the relevant range. Then, from this value, the square of the expectation (\(\mu^2\)) is subtracted. Following the provided steps, the variance for \(0 \leq x < 1\), was computed leading to \(\sigma^2 = \frac{1}{96}\), suggesting a very low spread around the expectation within the continuous range. As for the discrete part of the distribution, since there is no variability (the PDF is 0), the variance contribution is zero, reinforcing that the entirety of dispersion measured comes from the continuous portion of \(X\).