Show that adding a constant \(K\) to a random variable increases the average by \(K\) but does not change the variance. Show that multiplying a random variable by \(K\) multiplies both the average and the standard deviation by \(K\).

Short Answer

Expert verified
Adding \(K\) increases the mean by \(K\) but keeps the variance the same. Multiplying by \(K\) scales both mean and standard deviation by \(K\).

Step by step solution

01

Define the random variable and its properties

Let the random variable be denoted as \(X\) with the mean (expected value) \(E(X) = \mu_X\) and variance \(Var(X) = \sigma_X^2\). We'll consider two operations: adding a constant \(K\) and multiplying by a constant \(K\).
02

Adding a constant to the random variable

Consider the new random variable \(Y = X + K\). The mean of \(Y\) can be calculated as follows: \[E(Y) = E(X + K) = E(X) + K = \mu_X + K\]. Therefore, the mean of \(Y\) increases by \(K\).
03

Verify the variance when adding a constant

The variance of \(Y\) is calculated by: \[Var(Y) = Var(X + K) = Var(X) = \sigma_X^2\]. Constant \(K\) does not affect the variance.
04

Multiplying the random variable by a constant

Next, consider \(Z = KX\). The mean of \(Z\) is: \[E(Z) = E(KX) = K \times E(X) = K \times \mu_X\]. Hence, multiplying \(X\) by \(K\) multiplies its mean by \(K\).
05

Verify the variance when multiplying by a constant

The variance of \(Z\) is: \[Var(Z) = Var(KX) = K^2 \times Var(X) = K^2 \times \sigma_X^2\]. The standard deviation of \(Z\) is \sqrt{Var(Z)} = K \times \sigma_X\. Hence, multiplying \(X\) by \(K\) multiplies its standard deviation by \(K\).

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Key Concepts

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

mean of random variables
When dealing with random variables, understanding their mean (or expected value) is essential. The mean of a random variable provides a measure of its central tendency. It is similar to the average in a set of numbers and helps to identify the 'center' or 'typical' value around which the random variable tends to cluster. For a random variable \(X\), the mean is denoted as \(E(X)\) or \(\mu_X\).

Let's break it down:

- **Adding a Constant**: When you add a constant \(K\) to a random variable, the entire distribution shifts by that constant value. If \(Y = X + K\), then the mean of \(Y\) can be calculated as:

\[ E(Y) = E(X + K) = E(X) + K = \mu_X + K \]

This shows that the mean increases by the constant \(K\), but the shape of the distribution doesn't change.

- **Multiplying by a Constant**: When you multiply a random variable by a constant, the mean is scaled by that same constant. If \(Z = KX\), the mean of \(Z\) would be:

\[ E(Z) = E(KX) = K \times E(X) = K \times \mu_X \]

Thus, the mean is multiplied by \(K\). This operation scales all values of the random variable, but the 'center' of the distribution is adjusted accordingly.
variance of random variables
The variance of a random variable measures how spread out the values of the variable are around the mean. It gives us an idea about the variability or dispersion of the data. For a random variable \(X\), the variance is denoted as \(Var(X)\) or \(\sigma_X^2\).

Let's dive in:

- **Adding a Constant**: When a constant \(K\) is added to a random variable, the spread of the data does not change, only its position does. For \(Y = X + K\), the variance remains the same:

\[ Var(Y) = Var(X + K) = Var(X) = \sigma_X^2 \]

Thus, adding a constant does not change the variance.

- **Multiplying by a Constant**: When a random variable is multiplied by a constant, the variance is affected by the square of that constant. For \(Z = KX\), the variance of \(Z\) is given by:

\[ Var(Z) = Var(KX) = K^2 \times Var(X) = K^2 \times \sigma_X^2 \]

This means the variance is scaled by \(K^2\), indicating a larger spread if \(K > 1\) or a smaller spread if \(K < 1\).
standard deviation of random variables
The standard deviation of a random variable is the square root of its variance and offers a way of understanding the spread of the distribution in the same units as the random variable itself. For a random variable \(X\), the standard deviation is denoted as \(\sigma_X\).

Here's the breakdown:

- **Adding a Constant**: As with variance, adding a constant to a random variable does not change its standard deviation. For \(Y = X + K\):

\[ \sigma_Y = \sigma_X \]

The spread of the data remains the same; only the position of the data set shifts.

- **Multiplying by a Constant**: When multiplying a random variable by a constant, the standard deviation is scaled directly by that constant. For \(Z = KX\):

\[ \sigma_Z = K \times \sigma_X \]

Thus, the standard deviation is multiplied by \(K\), showing a direct scaling of the spread of the data based on the constant multiplier. This is particularly useful in fields like finance, where scaling risks and returns are common.

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Most popular questions from this chapter

Find the binomial probability for the given problem, and then compare the normal and the Poisson approximations. Find the probability of \(x\) successes in 100 Bernoulli trials with probability \(p=1 / 5\) of success (a) if \(x=25 ;\) (b) if \(x=21.\)

You are trying to find instrument \(A\) in a laboratory. Unfortunately, someone has put both instruments \(A\) and another kind (which we shall call \(B\) ) away in identical unmarked boxes mixed at random on a shelf. You know that the laboratory has \(3 A\) 's and \(7 B\) 's. If you take down one box, what is the probability that you get an \(A ?\) If it is a \(B\) and you put it on the table and take down another box, what is the probability that you get an \(A\) this time?

Let \(m_{1}, m_{2}, \cdots, m_{n}\) be a set of measurements, and define the values of \(x_{i}\) by \(x_{1}=\) \(m_{1}-a, x_{2}=m_{2}-a, \cdots, x_{n}=m_{n}-a,\) where \(a\) is some number (as yet unspecified, but the same for all \(x_{i}\) ). Show that in order to minimize \(\sum_{i=1}^{n} x_{i}^{2},\) we should choose \(a=(1 / n) \sum_{i=1}^{n} m_{i} .\) Hint: Differentiate \(\sum_{i=1}^{n} x_{i}^{2}\) with respect to \(a .\) You have shown that the arithmetic mean is the "best" average in the least squares sense, that is, that if the sum of the squares of the deviations of the measurements from their "average" is a minimum, the "average" is the arithmetic mean (rather than, say, the median or mode).

A shopping mall has four entrances, one on the North, one on the South, and two on the East. If you enter at random, shop and then exit at random, what is the probability that you enter and exit on the same side of the mall?

(a) Suppose you have two quarters and a dime in your left pocket and two dimes and three quarters in your right pocket. You select a pocket at random and from it a coin at random. What is the probability that it is a dime? (b) Let \(x\) be the amount of money you select. Find \(E(x)\) (c) Suppose you selected a dime in (a). What is the probability that it came from your right pocket? (d) Suppose you do not replace the dime, but select another coin which is also a dime. What is the probability that this second coin came from your right pocket?

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