How do we calculate the expected value E(X) for a density function?

In summary, E(X) or the expected value of a random variable is the sum of each value multiplied by its probability. For example, for a single dice roll, the expected value is 3.5. For three dice rolls, the expected value is much harder to calculate and will depend on the specific probabilities of each outcome. In the case of a continuous random variable with a density function, the expected value can be calculated using an integral. Additionally, the expectation operator has the property that it can be applied to constants and multiple random variables.
  • #1
EdmureTully
20
0
Give me an example like:

X: random variable representing the sum of 1 dice roll

p: probability of getting a dice result

E(X) = sum of X*p = 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) +6(1/6)

BONUS POINT:

what would be the E(X) for 3 dice rolls?

Sorry, I am really dumb. I am trying to understand exactly what E(X) means.
 
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  • #2
EdmureTully said:
what E(X) means.

It means the "expected value of X", which also called "the mean value of X". Are those terms defined in your study materials?
 
  • #3
I just don't really know how to find the E(X) of a density function and how to apply the theory for any case.
 
  • #4
Your situation isn't clear. Are you studying a book? Or are you trying to take some sort of quiz without having ever studied a book on probability theory or statistics? Your post which begins "Give me an example like" isn't even clear. Are you saying that someone gave you an example like that? Are you requesting another example?
 
  • #5
I am requesting an example for a density function and not a mass function.
 
  • #6
The problem, as you stated it, tells you all of that. For a single dice roll, there are 6 possible outcomes, all equally likely so the "density function" is f(1)= 1/6, f(2)= 1/6, f(3)= 1/6, f(4)= 1/6, f(5)= 1/6, f(6)= 1/6. The "mass function" (also called "cumulative probability function") is F(1)= 1/6, F(2)= 2/6= 1/3, f(3)= 3/6= 1/3, f(4)= 4/6= 2/3, f(5)= 5/6, f(6)= 6/6= 1.

E(x), the "expected value", is the sum of each value times it probability: 1(1/6)+ 2(1/6)+ 3(1/6)+ 4(1/6)+ 5(1/6)+ 6(1/6)= (1/6)(1+ 2+ 3+ 4+ 5+ 6)= (1/6)(21)= 3.5.

Of course, the calculations for three rolls are much harder. The lowest possible sum of three rolls is 1+ 1+ 1= 3 and that will occur only if all three rolls are 1. The probabability of that is (1/6)(1/6)(1/6)= 1/216. You can get a total of 4 with 2+ 1+ 1- that is 2 on die 1, and 1 on the other 2. Since the probability of anyone number is 1/6, the probability of "2, 1, 1" is also (1/6)(1/6)(1/6)= 1/216. But we can also get 4 with 1+ 2+ 1- 1 on die 1, 2 on die 2, and 1 on die 3, or 1+ 1+2- 1 on dice 1 and 2, 2 on die 3. The probability of each of those is each 1/216 but the probability of a total of 4 in any of those ways is 1/216+ 1/216+ 1/216= 3/216= 1/72. And for a total of 5, it gets worse! 1+ 1+ 3, 1+ 3+ 1, 3+ 1+ 1, 2+ 2+ 1, 2+ 1+ 2, 1+ 2+ 2, etc.
 
  • #7
EdmureTully said:
I am requesting an example for a density function and not a mass function.

In many books the term "density function" is often used for both discrete and continuous random variables, but perhaps your study materials use "mass function" exclusively for discrete random variables and you are asking for an example of computing [itex] E(X) [/itex] for a continuous random variable. Is that your question?
 
  • #8
EdmureTully said:
I am requesting an example for a density function and not a mass function.
The expected value for a real-valued random variable which has a density function ##p## is defined as
$$E[X] = \int_{-\infty}^{\infty} x p(x) dx$$
For example, if ##X## is uniformly distributed over the interval ##[2,4]##, then we have
$$p(x) = \begin{cases}
\frac{1}{2} & \text{ if }2 \leq x \leq 4 \\
0 & \text{ otherwise} \\
\end{cases}$$
and therefore
$$E[X] = \int_{2}^{4} \frac{1}{2} x dx = 3$$
 
  • #9
Also recall that the expectation operator has the following property:

E[aX+bY] = aE[X] + bE[Y] for constants a,b and random variables X and Y.
 

1. What is the formula for calculating the expected value E(X) for a density function?

The formula for calculating the expected value E(X) for a density function is: E(X) = ∫xf(x)dx, where x represents the random variable and f(x) represents the probability density function.

2. How do we interpret the expected value E(X) for a density function?

The expected value E(X) for a density function represents the mean or average value of the random variable X. It is the value that we would expect to obtain if we were to repeat the experiment multiple times.

3. What is the difference between the expected value E(X) and the actual value of X for a specific observation?

The expected value E(X) is a theoretical value based on the probability distribution of X, while the actual value of X is the value obtained from a specific observation. The expected value represents the long-term average, while the actual value can vary from observation to observation.

4. Can the expected value E(X) be negative?

Yes, the expected value E(X) can be negative. This can happen if the probability density function has a negative portion or if the distribution is skewed towards the left.

5. How does the expected value E(X) change if the probability distribution of X is skewed?

If the probability distribution of X is skewed, the expected value E(X) will be pulled towards the direction of the longer tail. This means that if the distribution is skewed towards the right, the expected value will be larger than the mean, and if the distribution is skewed towards the left, the expected value will be smaller than the mean.

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