I’m not sure if I understand what you mean by “expected value” (EVP). I don’t think it is a good idea to go for the “unexpected value” because that’s the number of “me” that the random variable is supposed to represent. The value is not the number of “me” that’s represented.

My mind works in a random-and-unpredictable way: I have a binomial distribution with n trials. It should be the number of me that the random variable is supposed to represent. If I look at the random variables inside of the binomial distribution and then I want to know what random variable is actually represented by, then I should go for the unexpected value.

The expected value of a binomial distribution is just the probability that this particular random variable actually happened to occur. In general, if I have n random variables, the probability of each of them happening is p = p(x_i), where x_i is the random variable. So the expected value is the probability that the random variable in question happened, which is simply p = p(x_i).

Binomial distributions are a good example of a random variable that has been assigned its expected value. It means that the probability of the random variable occurring is equal to the probability of the random variable happening. If we want to know the expected value of a binomial distribution, we can just look at the probability of each occurrence, which is p px_i.

For example, p x_i.Normal distribution is a distribution that contains two variables, like our previous example, but instead of a single variable there are many more. In particular, there is a “mean” variable that has a value, and a “standard deviation” parameter that controls how much the value of the mean varies from the value of the mean. For example, we can take a mean of 100 and assign a standard deviation of 10.

With our normal distribution, we would assign a probability of px_i of being in the x_ith bin of the distribution. The probability of the observed value being in the x_ith bin is p x_i. So what does it mean for the binomial distribution to have a binomial number of trials? Well, that’s exactly what the expected value of a binomial distribution with n trials tells you.

A binomial distribution with n trials, in which we have n observations, will have an expected value of the expected value of the binomial distribution.

It doesn’t say if the expected value of a binomial distribution with n trials is greater than the expected value of the binomial distribution with n trials. But if the expected value of a binomial distribution with n trials is greater than the expected value of the binomial distribution with n trials, then we’ll have a binomial distribution with the expected value of the binomial distribution with n trials.

I’m sure you get the point I’m trying to make here, but we have to remember the binomial distribution with n trials is the same thing as the binomial distribution with n trials. So the fact that the expected value of a binomial distribution with n trials is greater than the expected value of the binomial distribution with n trials. is a statement about the binomial distribution with n trials.

Well, if you take out a total of n trials in a binomial distribution where n trials equals 0, then the expected value of the binomial distribution with n trials is 0. That’s a statement about the binomial distribution with n trials.