If #X# is a random variable which can only assume the value #x#, then you have
#\mathbb{E}(X)=\mu = x#
This makes sense, since #X# assumes only one value, and so "on average" it assumes that value as well. Think for example that #X# equals constantly #3#. Then you sample, for example, ten values from #X#, and you will have #x_1 = 3#, #x_2 = 3, ... , x_10=3#, since you can't have anything but #3#. Now, compute the average:
#\mu = \frac{x_1 + ... + x_10}{10} = \frac{3+...+3}{10} = \frac{10\cdot 3}{10} = 3#
To be more precise, you may use the definition
#\mathbb{E}(X) = \sum p_i x_i#
i.e. the weighted sum of all possible values, weighted with their probabilities. Since #X# only assumes the value #x# with probability #1#, you have
#\mathbb{E}(X) = \sum p_i x_i = 1\cdot x = x#
