493. Find the point estimate of the unknown parameter $\lambda$ of the exponential distribution, with density $f(x)=\lambda \mathrm{e}^{-\lambda x}(x \geqslant 0)$, using the method of maximum likelihood based on the sample $x_{1}, x_{2}, \ldots, x_{n}$.
Solution. Let's form the likelihood function

$$
L=f\left(x_{1} ; \theta\right) \cdot f\left(x_{2} ; \theta\right) \ldots f\left(x_{n} ; \theta\right)
$$

considering that $\Theta=\lambda$ and, therefore, $f(x ; \theta)=f(x ; \lambda)=\lambda \mathrm{e}^{-\lambda x}$ :

$$
L=\left(\lambda \mathrm{e}^{-\lambda x_{1}}\right) \cdot\left(\lambda \mathrm{e}^{-\lambda x_{2}}\right) \ldots\left(\lambda \mathrm{e}^{-\lambda x_{n}}\right)=\lambda^{n} \cdot \mathrm{e}^{-\lambda \sum x_{i}}
$$

Let's find the logarithmic likelihood function:

$$
\ln L=n \ln \lambda-\lambda \sum x_{i}
$$

Let's find the first derivative with respect to $\lambda$ :

$$
\frac{d \ln L}{d \lambda}=n / \lambda-\sum x_{i}
$$

Let's write the likelihood equation by setting the first derivative to zero: $n / \lambda-\sum x_{i}=0$. Let's find the critical point by solving the obtained equation for $\lambda$ :

$$
\lambda=\frac{1}{\left(\sum x_{i}\right) / n}=1 / \bar{x}_{\mathbf{B}}
$$

Let's find the second derivative with respect to $\lambda$ :

$$
\frac{\mathrm{d}^{2} \ln L}{\mathrm{~d} \lambda^{2}}=(-n) / \lambda^{2}
$$

It is easy to see that at $\lambda=1 / \bar{x}_{\text {B }}$ the second derivative is negative; therefore, this point is a maximum point, and thus, the maximum likelihood estimate should be the reciprocal of the sample mean: $\lambda^{*}=1 / \bar{x}_{\text {B }}$.