Which is the expected value of the random variable with given probability distribution ?

Step-by-step explanation:

Let X be the RV we are talking about.

Expected value has different aliases, such as weighted average, mean, expectation, first moment and so on.

Okay, how do we calculate the average first, just sum all the values and divide by number of values. It comes from the basic assumption, which is choosing one of them is equally likely. What I mean by that is quite simple, above process what I talked about is just sum of the numbers times their probabilities. e.g. (a+b+c+d)/4 = a/4 + b/4 + c/4 + d/4. In the example, we assumed that probability of choosing any a, b, c and d is equally likely.

Then, why I use the term weighted average is the generalized situation of average, which is some of the numbers are repetitive. For instance, X can take the values a, b, c and d; however, X can take the value a 3 times more than d, X can take the value b 2 times more than d and c and d occurs with the same frequency at X, Therefore, if we take a sample, it is likely to be a, b, a, c, d, a, b or the permutation of this sample. Therefore the weighted average is 3 * a / 7 + 2 * b / 7 + c / 7 + d /7 or using normal average theorem (a + a + a + b + b + c +d) / 7. First, both of them are equal. Find the distribution and multiply the values with the probabilities of the distribution and sum them up or just add them and divide by number of sample. Second, if you check that, a has more weight, which means a can determine the direction of the average more. That is the logic behind the expected value.

Now, if X is a discrete RV (like in the above example), we know what we are going to do. What if X is a continuous RV ? X could take infinite values therefore if you divide a number with an infinite value, the probability of an individual value X could take is going to be 0. Therefore, we cannot calculate the expected value of chosen values individually; however, we can calculate the expected value of an interval by approximating with Riemann Sum or with a better choice, using integrals. Because, technically using Riemann Sum and quantizing the values and approximating the area of the curve is the same as we did above.

In conclusion, basically if X is a discrete RV and the values X could take is countably infinite,

If X is a discrete RV and the values X could take is finite,

if X is a continuous RV,

the answer is 13

step-by-step explanation:

5*5=25

12*12=144

25+144=169

169^2=13