biasedness/unbiasedness of an MLE. To show whether an MLE I just found is biased/unbiased, would I need to find the expectation of the answer? Plus would I do this by integrating $\text{MLE}  \cdot \text{pdf}$. 
My MLE is $ \frac{1}{\bar x} $
I've heard the expectation of this is the same as of the expectation of $ \frac{1}{x} $
http://www2.imperial.ac.uk/~ayoung/m2s1/Exercises8.PDF question 6 part 2, I differentiated th log likelihood and set to zero to get $ \hat\theta =  \frac{n}{sum...} = \frac{1}{\bar x} $
 A: I take it that you are dealing with the exponential distribution with
$$
    f_X(x) = \lambda \mathrm{e}^{-\lambda x} [x > 0]
$$
Assuming all elements of the sample $\{x_1,x_2,\ldots,x_n\}$ are positive, the log-likelihood reads:
$$
  n \log \lambda - \lambda \sum_{k=1}^n x_k
$$
which attains its maximum exactly at $\lambda = \frac{n}{\sum_{k=1}^n x_k}$.
Now to computation of the expectation of the MLE: 
$$\begin{eqnarray}
  \mathbb{E}\left(\frac{n}{X_1+X_2+\cdots+X_n}\right) &=& n \mathbb{E}\left(\frac{1}{X_1+X_2+\cdots+X_n}\right) \\ &=& n \mathbb{E}\left( \int_0^\infty \exp\left(-t(X_1+\cdots+X_n)\right) \mathrm{d}t \right) \\ &=& n \int_0^\infty \mathbb{E}\left(  \exp\left(-t(X_1+\cdots+X_n)\right)  \right) \mathrm{d}t  \\ &\stackrel{\text{indep.}}{=}& n \int_0^\infty \left(\mathbb{E}\left(  \exp\left(-t  X_1\right)  \right)\right)^n \mathrm{d}t \\ &=&
 n \int_0^\infty \left(\frac{\lambda}{t+\lambda}\right)^n \mathrm{d}t = \left. -\frac{n}{n-1} \frac{\lambda^n}{(t+\lambda)^{n-1}} \right|_{0}^\infty \\
  &=& \frac{n}{n-1} \lambda
\end{eqnarray}
$$
Thus the MLE is biased.
