# Is my understanding of Expected Value of a Random Variable correct?

I have come up with the following understanding about Expectation of RV.

Please, correct me if I am wrong.

Definition: Expected value of a RV says that, if a random experiment is repeated n number of times, what would be the value that you expect to see most of the times. Expected value of a Random Variable is a value. It is not a probability.

Example: In case of a dice, if the dice is rolled $n$ number of times, the value we would see most of the times is 3.5.

Calculation: We don't have to repeat the same experiment $n$ number of times to find the expected value of a random variable. It can be done using the random variable and its associated probability distribution. In case of a dice,

$\sum_{x=1}^6 x(1/6) = 1*(1/6) + 2*(1/6) + 3*(1/6) + 4*(1/6) + 5*(1/6) + 6*(1/6) = 3.5$

Continuous Case

If $X$ is a continuous random variable with density function $f_X(x),$ then the expectation of $X$ is $\displaystyle E(X)=\int_{-\infty}^\infty x\,f_X(x)\,dx$

If $Y=g(X)$ is a function of $X$(i.e. a density function)$,$ then $\displaystyle E(Y)=\int_{-\infty}^\infty g(x)\,f_X(x)\,dx$

Edit:

After reading comments and answers,

Definition:

Expected value of a RV says that, if a random experiment is repeated n number of times, what would be the average value of the outcomes.

Expected value of a Random Variable is a value. It is not a probability.

Example:

In case of a dice, if the dice is rolled $n$ number of times, the average value of the outcomes would be 3.5.

• This definition is wrong, it defines mode, not EV. Consider a fair coin (0 - tails, 1 - heads), it's expected value is 1/2 * 0 + 1/2 * 1 = 0.5. You will never see value of 0.5 since this variable has only values in {0, 1}. – Tim Oct 6 '16 at 12:58
• @Tim, How can I correct my definition in its current form? – user366312 Oct 6 '16 at 13:01
• I, for one, have never seen a value of 3.5 on a die. – whuber Oct 6 '16 at 13:08
• @whuber, Is my edit correct now? – user366312 Oct 6 '16 at 13:13
• Your edit, which reflects the answer you have received, expresses a derived result: it is a (qualitative) statement of the weak law of large numbers. Although that certainly can help with understanding, it is not a definition of expected value. Your original "example" still expresses the bizarre idea that a die can actually exhibit a value of 3.5! The expected value is a property of a model for your experiment (a die roll), as I have explained in various answers such as the one at stats.stackexchange.com/a/30330/919. – whuber Oct 6 '16 at 13:16

## 2 Answers

The expected value can be thought of as the long-run average. If you roll your die many, many times, and look at the average of the rolls, you will end up close to 3.5.

This is not the value that you expect to see most often. This would be the mode of the distribution (in the discrete case - for a continuous distribution, any particular value has the same probability of zero). After all, you will never see a roll of 3.5 with your standard six-sided die.

I would like to help - you put $n$ in your definition, but this is exactly what we don't want. Since $n$ is never referenced elsewhere, its presence serves to inform the reader of your definition that the experiment is repeated a finite number of times. So, we can either remove $n$ altogether and state that the experiment is repeated infinitely many times (which is a problem logically) or we can leave $n$ right where it is and state that the expectation is the limit of the average of the values as $n\to\infty$. That is, in your dice example, let's take the average of $n$ values. As we increase $n$, the average gets closer and closer to $3.5$.

• This is a statement of a law of large numbers, but it does not define the expectation (which seems to be the focus of your interest). No mathematical definition of expectation employs a limit over an infinite sequence of experiments. For discrete finite random variables, like the one describing the roll of a die, the definition is a finite sum: no limits are needed at all. – whuber Sep 28 '17 at 15:06
The text passage contains a definition and explanation of the expected value of a random variable, including an example with a die and the calculation of the expected value. It also includes a discussion about the continuous case and a correction to the initial definition based on feedback.
no_question

the passage does not contain a direct mathematical question that requires an answer. it is more of an explanation and a discussion of the concept of expected value. however, it does contain a calculation of the expected value of a die roll, which is 3.5, but this is not presented as a question.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Is my understanding of Expected Value of a Random Variable correct?

I have come up with the following understanding about Expectation of RV.

Please, correct me if I am wrong.

Definition: Expected value of a RV says that, if a random experiment is repeated n number of times, what would be the value that you expect to see most of the times. Expected value of a Random Variable is a value. It is not a probability.

Example: In case of a dice, if the dice is rolled $n$ number of times, the value we would see most of the times is 3.5.

Calculation: We don't have to repeat the same experiment $n$ number of times to find the expected value of a random variable. It can be done using the random variable and its associated probability distribution. In case of a dice,

$\sum_{x=1}^6 x(1/6) = 1*(1/6) + 2*(1/6) + 3*(1/6) + 4*(1/6) + 5*(1/6) + 6*(1/6) = 3.5$

Continuous Case

If $X$ is a continuous random variable with density function $f_X(x),$ then the expectation of $X$ is $\displaystyle E(X)=\int_{-\infty}^\infty x\,f_X(x)\,dx$

If $Y=g(X)$ is a function of $X$(i.e. a density function)$,$ then $\displaystyle E(Y)=\int_{-\infty}^\infty g(x)\,f_X(x)\,dx$

Edit:

After reading comments and answers,

Definition:

Expected value of a RV says that, if a random experiment is repeated n number of times, what would be the average value of the outcomes.

Expected value of a Random Variable is a value. It is not a probability.

Example:

In case of a dice, if the dice is rolled $n$ number of times, the average value of the outcomes would be 3.5.

• This definition is wrong, it defines mode, not EV. Consider a fair coin (0 - tails, 1 - heads), it's expected value is 1/2 * 0 + 1/2 * 1 = 0.5. You will never see value of 0.5 since this variable has only values in {0, 1}. – Tim Oct 6 '16 at 12:58
• @Tim, How can I correct my definition in its current form? – user366312 Oct 6 '16 at 13:01
• I, for one, have never seen a value of 3.5 on a die. – whuber Oct 6 '16 at 13:08
• @whuber, Is my edit correct now? – user366312 Oct 6 '16 at 13:13
• Your edit, which reflects the answer you have received, expresses a derived result: it is a (qualitative) statement of the weak law of large numbers. Although that certainly can help with understanding, it is not a definition of expected value. Your original "example" still expresses the bizarre idea that a die can actually exhibit a value of 3.5! The expected value is a property of a model for your experiment (a die roll), as I have explained in various answers such as the one at stats.stackexchange.com/a/30330/919. – whuber Oct 6 '16 at 13:16

## 2 Answers

The expected value can be thought of as the long-run average. If you roll your die many, many times, and look at the average of the rolls, you will end up close to 3.5.

This is not the value that you expect to see most often. This would be the mode of the distribution (in the discrete case - for a continuous distribution, any particular value has the same probability of zero). After all, you will never see a roll of 3.5 with your standard six-sided die.

I would like to help - you put $n$ in your definition, but this is exactly what we don't want. Since $n$ is never referenced elsewhere, its presence serves to inform the reader of your definition that the experiment is repeated a finite number of times. So, we can either remove $n$ altogether and state that the experiment is repeated infinitely many times (which is a problem logically) or we can leave $n$ right where it is and state that the expectation is the limit of the average of the values as $n\to\infty$. That is, in your dice example, let's take the average of $n$ values. As we increase $n$, the average gets closer and closer to $3.5$.

• This is a statement of a law of large numbers, but it does not define the expectation (which seems to be the focus of your interest). No mathematical definition of expectation employs a limit over an infinite sequence of experiments. For discrete finite random variables, like the one describing the roll of a die, the definition is a finite sum: no limits are needed at all. – whuber Sep 28 '17 at 15:06
user
The text passage contains a definition and explanation of the expected value of a random variable, including an example with a die and the calculation of the expected value. It also includes a discussion about the continuous case and a correction to the initial definition based on feedback.

TYPE: NO_QUESTION

The passage does not contain a direct mathematical question that requires an answer. It is more of an explanation and a discussion of the concept of expected value. However, it does contain a calculation of the expected value of a die roll, which is 3.5, but this is not presented as a question.