(See page 50 of the Cryer and Chan text, Exercise 3.2)

Y-bar = (1/n)*sum_{(from t=1 to t=n)}(Y_t).  For our fifth process, this becomes: Y-bar = (1/50)*sum_{(from t=1 to t=50)}(e_t-e_t-1).  We remove mu because we know the variance of a constant is zero. 

The summation yields: (e₁-e₀)+(e₂-e₁)+(e₃-e₂)+...+(e₄₉-e₄₈)+(e₅₀-e₄₉).  Note the cancelations within. 

Plugging this summation back in, we find Y-bar = (1/50)(e₅₀-e₀).  From here we can easily find the variance: 

Var(Y-bar) = (1/50)²Var(e₅₀-e₀).  Recall, the e's are independent and identically distributed, so you should reach your original 2/2500.  Let me know if you have any further questions.

RDH