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Chapter 4: Problem 6

If \(X\) is a random variable, explain whether it is true that \(X+X=2 X\) and \(X-X=0\). Are 0 and \(2 X\) random variables?

Short Answer

Expert verified
Yes, 0 is not a random variable but 2X is.

Step by step solution

01

Understand the Problem

We need to verify if the equations \(X + X = 2X\) and \(X - X = 0\) hold true for random variables, and determine if 0 and \(2X\) are random variables.
02

Consider Properties of Random Variables

A random variable is a numerical outcome of a random phenomenon. It's important to remember that basic arithmetic operations on a random variable follow the same algebraic rules as numbers.
03

Evaluate X + X = 2X

According to the distributive property in algebra, \(X + X\) simplifies to \(2X\). Thus, for a random variable \(X\), \(X + X = 2X\) is true.
04

Evaluate X - X = 0

Subtraction of any value by itself results in zero. Therefore, for a random variable \(X\), \(X - X = 0\) holds true as well.
05

Determine if 0 is a Random Variable

Zero itself is not random; it represents a constant outcome with no variability. Hence, 0 is not a random variable.
06

Determine if 2X is a Random Variable

When a random variable \(X\) is multiplied by a constant, like 2, the result \(2X\) is still a random variable because it is dependent on the stochastic nature of \(X\).

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Key Concepts

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

Arithmetic Operations
Arithmetic operations, such as addition and subtraction, play a significant role in the manipulation of random variables. These operations on random variables abide by the same rules and laws that apply to regular numbers.
For example, when you add a random variable X to itself, effectively performing the operation \(X + X\), you are still applying the basic arithmetic operation of addition, resulting in \(2X\). This is akin to how it would work with any regular number. Similarly, subtracting a random variable from itself, \(X - X\), naturally yields zero.
This adherence to arithmetic operations facilitates an easier manipulation of random variables, enabling their use in various statistical analyses and problem-solving scenarios.
Distributive Property
The distributive property is a fundamental principle in algebra that also applies to random variables. It states that multiplying a number by a sum is the same as doing each multiplication separately, and then adding the results.
In the context of random variables, if you have \(X + X\), this can be considered as \((1 + 1)X\), which directly simplifies to \(2X\) through distributive property. It demonstrates that combining terms of a random variable follows the same logic as with other algebraic expressions.
Understanding this property can help in simplifying complex expressions involving random variables and is crucial in ensuring that mathematical operations on these variables yield accurate results.
Constant Outcome
A constant outcome refers to a value that does not change. Within the realm of probability and random variables, a constant outcome signifies a result with no randomness involved.
For instance, zero is a constant outcome. When you subtract a random variable from itself, \(X - X = 0\), the result zero doesn't embody any unpredictable element 鈥� it's simply constant.
Consequently, constant outcomes cannot be considered random variables as they lack variability and randomness, central characteristics that define random variables.
Stochastic Nature
The stochastic nature of a random variable refers to its inherent randomness and variability. This nature is what characterizes random variables and differentiates them from fixed numbers.
When a random variable \(X\) is involved in an operation, such as being multiplied by a constant, the result, like \(2X\), retains this stochastic quality. Although \(2X\) implies the outcome is scaled, it is still contingent on the unpredictable nature of \(X\).
Thus, any manipulation or transformation of a random variable continues to embody its stochastic essence unless explicitly made deterministic, as is the case with constant outcomes.

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Most popular questions from this chapter

Let \(\left(x_{i} ; 1 \leq i \leq n\right)\) be a collection of positive numbers. Show that $$ \left(\frac{1}{n} \sum_{i=1}^{n} \frac{1}{x_{i}}\right)^{-1} \leq\left(\prod_{i=1}^{n} x_{i}\right)^{1 / n} . $$

Let \(f_{1}(X)\) and \(f_{2}(X)\) be functions of the random variable \(X\). Show that (when both sides exist) \(\left[\mathbf{E}\left(f_{1} f_{2}\right)\right]^{2} \leq \mathbf{E}\left(f_{1}^{2}\right) \mathbf{E}\left(f_{2}^{2}\right)\). Deduce that \(\mathbf{P}(X=0) \leq 1 \quad[\mathbf{E}(X)]^{2} / \mathbf{E}\left(X^{2}\right)\). (Recall that \(a t^{2}+2 b t+c\) has distinct real roots if and only if \(b^{2}>a c\).)

Cars are parked in a line in a parking lot in order of arrival and left there. There are two types of cars, small ones requiring only one unit of parking length (say \(15 \mathrm{ft}\) ) and large ones requiring two units of parking length (say \(30 \mathrm{ft}\) ). The probability that a large car turns up to park is \(p\) and the probability that a small car turns up is \(q=1-p\). It is required to find the expected maximum number of cars that can park in a parking length of \(n\) units, where \(n\) is an integer. Denoting this number by \(M(n)\) show that: (a) \(M(0)=0\) (b) \(M(1)=1-p\) (c) \(M(n)-q M(n-1)-p M(n-2)=1, \quad(n \geq 2)\) Show that the equations are satisfied by a solution of the form \(M(n)=A \alpha^{n}+B \beta^{n}+C n\), where \(\alpha, \beta\) are the roots of the equation \(x^{2}-q x-p=0\), and \(A, B, C\) are constants to be found. What happens to \(M(n)\) as \(n \rightarrow \infty\) ?

Preparatory to a camping trip, you can buy six cans of food, all the same size, two each of meat, vegetables, and fruit. Assuming that cans with the same contents have indistinguishable labels, in how many distinguishable ways can the cans be arranged in a row? On the trip, there is heavy rain and all the labels are washed off. Show that if you open three of the cans at random the chance that you will open one of each type is \(\frac{2}{5}\). If you do not succeed, you continue opening cans until you have one of each type; what is the expected number of open cans?

Let \(X \geq 0\) be integer valued. Use the indicator \(I(X>k)\) to prove that $$ \mathbf{E} X=\sum_{k \geq 0} \mathbf{P}(X>k) $$ and $$ \mathbf{E} X^{r}=\sum_{k \geq 0} r k^{r-1} \mathbf{P}(X>k) $$
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