Expected Value of a Random Variable

Transcription

1 Knowledge Article: Probability and Statistics Expected Value of a Random Variable Expected Value of a Discrete Random Variable You're familiar with a simple mean, or average, of a set. The mean value of a random variable is somewhat different. It does not make sense to find the simple average of the values of a random variable because each value may come with a different probability. Think of a loaded die with a 50% chance of rolling a 6 and a 10% chance of rolling each of the other numbers. If you were to roll such a die 10 times, you'd expect 6 to turn up 5 times (as shown in the graph). This is in agreement with in fact, follows from the definition of probability, which is the frequency of occurrence of an event in a situation of chance. So it seems reasonable that the mean of the random variable should be closer to 6 than the simple average, which is 3.5. (This is calculated by multiplying ) That is indeed the case; here, the mean works out to 4.5. This is the expected value for this die. 0.5 P(X = x i ) Probability Distribution of X mean (X) = x i 1

2 Expected value is calculated by multiplying each outcome of the random variable X (the top face of the die) by the probability that it can occur. The sum of the products is the expected value. Here, = 4.5. In general, we can say that expected value is found by multiplying the probability of each outcome by the payoff. The payoff in this case is as simple as the number of the dots on the upper face of the die. The expected value is sometimes referred to as a weighted average. It is important to note that the mean of a discrete random variable is not necessarily one of the allowed values of the variable. Clearly, 4.5 will never appear on any face of the die! Probability Distributions We can also take data from a normal distribution to a standard normal curve, where we know the probabilities. The payoffs will be associated with a range of outcomes, rather than individual outcomes. So once again, we can calculate the expected values by multiplying the probability by the payoff and adding any separate terms. Consider this situation. The mean score on a widely used IQ (intelligence quotient) test is 100, and the standard deviation is 15 points. We can calculate a new standard deviation of unit 1 by finding the z-score, or the number of deviations a value is from the standard mean of 0. The formula is z = x μ, where x is the random variable. What is σ the probability of having an IQ score of 115 or above? z = = The z-score is 1, meaning one deviation above the mean. Using the rule to estimate, we have 50% + 34%, so the remainder, %, or 16%, is the probability that a person selected at random from the population has an IQ greater than 115. Essentially, we move the initial normal curve so it is symmetric around 0 and then adjust the deviation to a standard unit of 1, 2, 3, etc. What is the probability of having a score of 70? z = , 15 or -2. Since z = -2 means two deviations below the mean, there is a 2.5% chance of choosing a random person whose IQ is below 70, according to the rule. 2

3 Mean ( ) The mean, represented by, specifies the center of a normal curve. Note these three elements of the curve: Symmetry (1). The curve of a normal distribution is symmetric about the axis x =. The parts of the curve to the left and to the right of this axis are mirror images of each other. This property can be represented by the equation P( μ x) P( μ x). Peak (2). The highest point of the normal curve occurs at the mean value. This means that the probability density is greatest at the center. Probabilities (3). The areas under the curve to the left of the mean and to the right of the mean are both 50% of the entire area under the curve. This means a normal variable has equal probabilities of being greater than or smaller than the mean. This property reflects the symmetry of the normal curve. Standard Deviation ( ) The standard deviation, represented by, determines the height and width of a normal curve. 3

4 Note these three elements of the curve: Height (1). The height of the peak of the normal curve depends on the value of the standard deviation ( ). A greater value of implies a lower peak height. Width (2). The width of the peak of the normal curve at any fraction of the maximum height depends on the value of the standard deviation. A greater value of gives a bigger width, or spread, of the distribution. Probability density (3). Most of the area under the normal curve lies within a few standard deviations of the mean. Standard Normal Distribution The normal curve with a mean of 0 and a standard deviation of 1 is known as the standard normal distribution. This curve is easier to work with for the purposes of computation: it can be shown that any normal curve can be mapped to the standard normal curve. Note these three characteristics of this graph: Mean (1). The mean of a standard normal distribution is = 0. The peak (center) of the distribution occurs at x = 0. Symmetry (2). The standard normal distribution is symmetric about the probability axis x = 0. This means the density function has the property P(-x) = P(x). Standard deviation (3). The standard normal curve has a standard deviation of exactly 1: = 1. Probabilities That Do Not Follow the Rule What if the standard deviation does not come out to a whole number? The probability associated with any z-value can be calculated using integral calculus, but most analysts use electronic tools to get these probabilities. We will use the same tools in this course. Taking calculus will give you a better understanding of how these tools actually work. Let s look at an example. Suppose someone has an IQ of 120. What is the probability that another randomly chosen person will have a higher IQ? 4

5 Draw a normal curve and mark the mean and standard deviations on it. You can tell that 120 is more than one standard deviation above the mean. In fact, it is z = This is 1.33 deviations above the mean. Using an electronic tool, we see that the probability up to 120 is So the probability of another randomly chosen person having an IQ above that is , or Therefore, in a randomly chosen group of 100 people, you could expect to find that about 9% of people have an IQ above 120. The number of people out of 100 would be considered the expected value in this case. The expected value is about 9 people, or more precisely, 9.1 people, because the expected value does not have to be an actual value that could occur. Normal curve This knowledge article is adapted from the following sources: PLATO Precalculus lesson Discrete Random Variables 5