by llamaro25
Last Updated September 12, 2019 03:19 AM

The exponential distribution is a probability distribution often used to model waiting or survival times. The version that we will look at has a probability density function of the form

$$p(y | v) = exp((-e^-v)y - v)$$

where $0<y<\infty$, i.e., y can take on the values of non-negative real numbers. In this form it has one parameters: a log-scale parameter $v$. If a random variable follows a gamma distribution with log-scale $v$ we say that $Y ~Exp(v)$. If Y ~Exp(v), then $E[Y]= e^v$ and V[Y]= e^(2v).

The question is;

Determine the approximate bias of the maximum likelihood estimator $\hat{v}$ of $v$ for the exponential distribution.

The MLE that I have gotten is $\hat{v}= \ln (m/n)$ where $m=\sum_1^n y_i$

I know that the formula for bias is $b(\hat{v}) = E(\hat{v}) - v$

Here's my attempt to calculate the bias.

$$b(\hat{v}) = ln(m/n) - v$$ since $m/n$ is the sample mean, $m/n = E[Y]= E(m/n) = e^v$

This implies that

$b(\hat{v}) = ln(e^v)- v = 0$ So the bias is 0? I'm not sure if I calculated the MLE and bias correctly as I'm new to statistics

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