# Expected value of a random variable differing from arithmetic mean

I have seen that expected value of a discrete random variable is equal to the arithmetic mean of the distribution provided the values it takes. Is it true for all random variables irrespective of the distribution? Is there a case or example where expected value differs from the arithmetic mean?

Secondly I think it applies only for discrete random variables. I think for continuous random variables, the pdf is zero at particular points. So in that case can I say that expected value is not equal to the mean of random variable?

• Welcome to the list. Is this homework? If so, you should add the self-study tag Commented Oct 7, 2013 at 11:56
• For the discrete case when there are $n$ different values random variable $X$ can take, $EX=\sum_{i=1}^nx_iP(X=x_i)$, so unless $P(X=x_i)=1/n$ the expected value does not equal to the arithmetic mean. Commented Oct 7, 2013 at 12:20
• @PeterFlom No This is not homework. I had a doubt and wanted to clarify it. Commented Oct 7, 2013 at 12:55
• @mpiktas when you say P(X=Xi)=(1/n), isn't that uniform distribution. So can i understand it as only when the distribution is uniform the arithmetic mean is equal to the expected value. Commented Oct 7, 2013 at 12:59
• Yes, you can say that. The correct mathematical statement would be that the expected value of discrete variable with finite number of different values is equal to the arithmetic mean of those values only in the case when the discrete variable is uniform. Commented Oct 7, 2013 at 13:02

In the discrete case the expected value is a weighted sum, where the possible values of the variable are weighted by their probability of occurring (the probability mass function), $EX=\sum_{i=1}^nx_iP(X=x_i)$. Since all weights are non-negative, smaller than untiy, and their sum equals unity, the expected value of a discrete random variable is also a specific convex combination of its possible values.

In the continuous case the expected value is a weighted integral, where the possible values of the variable are weighted by the probability density function $EY=\int_{-\infty}^{\infty}yf_Y(y)dy$.

What happens is that the arithmetic (i.e. unweighted) mean from the realization of a collection of identically distributed random variables (i.e. the "sample mean") is shown to be an unbiased and consistent estimator of the expected value, although the latter is a weighted mean.

• very clearly explained. Does that mean that arithmetic mean is an unbiased estimator of the expected value given any distribution? Though I am not aware of what an unbiased estimator is? Commented Oct 7, 2013 at 13:29
• I would suggest to stop using the term "arithmetic mean" in order to refer to the "sample mean". Let me put it this way: the "sample mean" is an arithmetic mean. And yes the sample mean is an unbiased estimator of the expected value of any distribution, where "unbiased estimator" means, that the expected value of the estimator equals the true value that the estimator estimates. In our case, since the sample mean $\bar X = \frac 1n\sum_{i=1}^{n}X_i$ attempts to estimate the expected value $EX$ of the distribution that all $X_i$ follow, we have $E(\bar X) = E(X)$. Commented Oct 7, 2013 at 14:49
• You have used the term sample mean but then you have used arithmetic mean expression in it. what is the difference between them? Can you please elaborate on what does it mean by saying that the expected value of sample mean(in this case arithmetic mean) is equal to expected value of distribution that Xi follows. Commented Oct 7, 2013 at 17:21
• You need to review basic definitions. Start with the wikipedia article en.wikipedia.org/wiki/Mean_value about all different kinds of mean values. "Arithmetic mean" is one kind of mean value. "Sample mean" is just the "mean value of the sample", and it is an arithmetic mean. Commented Oct 7, 2013 at 22:09
• I checked the link. I think sample mean refers to a arithmetic mean of a finite no of samples taken in a distribution you choose and the expected value of this sample mean is an estimator of the expected value of the entire distribution(population mean).one small question is say you are taking less no of samples and calculating sample mean. Now there might be a lot of variation in distribution later that doesn't show up in these samples. so when you use the expected value of this sample mean as an estimate of total mean, does the distribution function expression take care of this? Commented Oct 8, 2013 at 3:19

I think that an arithmetic mean approaches expected value, as the number of samples increase in number. Say, you have a die which you have rolled 10 times and the outcomes are {5,6,4,5,3,2,1,2,4,6}

The mean of the above values is 3.8. But the expected value when a die is rolled 10 times ( for that matter any number of times) is constant and is 1*1/6+2*1/6+...6*1/6 = 3.5

Hence, we see that the mean and expected values are different. If you take a large number of samples, the mean of the sample means (population mean) will reach the expected vale