When can I not replace a random variable with its mean? A frequent simplification in modeling and simulation is to replace a random variable by its mean value. 
When would this simplification lead to the wrong conclusion?
 A: If you replace a missing value by some point estimate, you disregard all its variability. Thus, you will not propagate all the original variability to your model. Your parameter estimates will appear to have too low standard-errors. If you do inference, your p values will be biased low. Your confidence-intervals will be too narrow. If you do prediction, your prediction-intervals will be too narrow.
Overall: you will be too sure of your conclusions.
A: In addition to Stephan's points:


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*In almost any application where you're interested in nonlinear functions of the random variable, substituting the mean will generally introduce bias and possibly contradictory results. The average velocity and average mass of a particle will generally not be consistent with average kinetic energy, because energy scales with V^2.

*The mean value may not even be a possible outcome for the random variable. If my possible outcomes are 0 "patient dies" and 1 "patient lives", it's probably not helpful to have a model that describes the patient as 0.1 "mostly dead but slightly alive".

A: A real life example (related to the two answers you got), in the financial markets. The price of an option is based in the probability that the price of an asset goes above (or below) a given level.
For example, the price of an option for buying an asset at a price 100 when the expected value of the asset is 80. If you substitute the random variable (the asset price) by its mean, you would get a price of zero (as you would never by at 100 an asset that costs 80). When you take into account the stochasticity of the asset (and that's the right way of doing it) you get a positive price, as there is some probability that the asset price goes above 100.
