An exponential distribution has a probability density function: $$p(y|v) = exp(−e^{−v}y − v)$$
It has one parameter: a log-scale parameter $v$. If a random variable follows a gamma distribution with log-scale $v$ then $Y ∼ Exp(v)$. If $Y ∼ Exp(v)$, then $E[Y] = e^v$ and $V[Y] = e^{2v}$.
I have computed the likelihood function to be:
$$p(y_1,y_2,...,y_n|v) = e^{-(−e^{−v} \sum_{i=1}^{n} y_i + nv)}$$
Then taking the negative logarithm of the likelihood expression, the negative log likelihood is
$$L(y_1,y_2,...,y_n|v) = e^{-v} \sum_{i=1}^{n} y_i + nv$$
Next, I have to find the MLE $\hat{v}$ for $v$. This is the value of $v$ that minimises the negative log likelihood. I found $$\hat{v} = -\ln(\frac{n}{\sum_{i=1}^{n} y_i})$$
Using the above, I need to determine the bias and variance for the MLE. How do I go about doing that? Also do correct me if the above calculations are incorrect.