# Estimator for $\frac{1}{\lambda}$ using $\min_i X_i$ when $X_i$ are i.i.d $\mathsf{Exp}(\lambda)$

Let $$X_1,\ldots,X_n$$ be i.i.d. $$\mathsf{Exp}(\lambda)$$ random variables, where $$\lambda$$ is unknown. Consider $$f_{\min}(x) = \min_{i}(X_i)=$$ $$n \lambda$$ Exp$$(n\lambda x)$$.

I am told that $$\hat \theta \mathrel := n \cdot \min_i(X_i)$$ is an unbiased estimator for the parameter $$1/\lambda$$. Indeed, this is true since the expected value of the above defined $$\hat \theta$$ is equal to $$1/\lambda$$. But, in this setting how would one proceed to compute $$1/\lambda$$ from $$\hat \theta$$? (don't we construct an estimator in order to compute a parameter?)

...in this setting how would one proceed to compute $$1/\lambda$$ from $$\hat \theta$$?

You can't compute $$1/\lambda$$ because $$\lambda$$ is the unknown parameter. That is why we estimate it. We can take the value $$\hat{\theta}(\mathbf{x}) = n x_{(1)}$$ to be an estimate of $$1/\lambda$$ (and we can compute that estimate from the observed data), but we will not know the true value of the parameter.

• thanks for commenting. Indeed, that was I meant, e.g. how to compute an estimate of it. I thnk I got confused thinking about cases where we use the sample mean as an estimator. I see it more clearly now. By the way, $\hat \theta$ has a quadratic risk of $1/\lambda^2$. Since it's unbiased is it also consistent? Mar 4, 2019 at 12:25

You know that $$\hat\theta(X_1,\ldots,X_n)=n\min_\limits{1\le i\le n} X_i$$ is an unbiased estimator of $$\frac{1}{\lambda}$$, a function of the unknown parameter $$\lambda$$.

This implies that an unbiased estimate of $$\frac{1}{\lambda}$$ is given by $$\hat\theta(x_1,\ldots,x_n)=n\min_\limits{1\le i\le n} x_i$$. [Here $$x_i$$ denotes the observed value of $$X_i$$ in the sample]. This is the value of $$\frac{1}{\lambda}$$ you are seeking based on the observed sample $$x_1,\ldots,x_n$$.

• I'm not sure I understand the distinction you're trying to make here. The concept of "bias" applies only to an estimator, not an estimate. Can you clarify?
– Ben
Mar 4, 2019 at 11:01
• I am not making any distinction; it is just a way of saying that $\frac{1}{\lambda}$ is unbiasedly estimated by $\hat\theta$. Mar 4, 2019 at 11:21