Find $r$ when $r>15$ A light-bulb flickers after every so often in minutes, and the time taken between each flicker is recorded. In total 5 observations are counted, and the time between each has been recorded as the following:
$$x_1=2.5, x_2=5.4, x_3 = 6.4, x_4 = 2.1$$
However, the 5th observation is only recorded when the time between flickers is greater than 15 minutes, so we have $x_5 > 15$.
Calculate the sample mean.
Here's what I have tried:
$$\frac{1}{n}\sum_{i=1}^5x_i = \frac{2.5+5.3+6.4+2.1+r}{5}=3.26+\frac{r}{5}$$
Where $r$ represents $x_5 > 15$. However, how do I find a value for $r$ if possible? The original question asks for this distribution to find the MLE of $X \sim \exp(\lambda;x)$. Give the MLE of $\exp(\lambda;x)$ is $\bar{x}$, I thought the interpretation I gave would be the answer. Please let me know if an alternative approach was required!
 A: I assume $X_1,X_2,\ldots,X_5$ are i.i.d Exponential random variables with mean $1/\lambda$. Let $f$ be their common density function. Regarding your original question on MLE of $\lambda$, I think what you have is an instance of Type-I (right) censoring. There are $4$ uncensored observations and the $5$th observation is right-censored.
So for $x_i>0$ and $\lambda>0$, the likelihood here takes the form
\begin{align}
L(\lambda \mid \boldsymbol x)&=\prod_{i=1}^4f(x_i)\cdot P(X_5>15)
\\&=\prod_{i=1}^4 (\lambda e^{-\lambda x_i})\cdot e^{-15\lambda}
\\&=\lambda^4 \exp\left\{-\lambda\sum_{i=1}^4 x_i-15\lambda\right\} \tag{1}
\end{align}
One can also say that the data is $(Y_i,\delta_i)$ where $Y_i=\min(X_i,15)$ and $\delta_i=I(X_i\le 15)$, for which the likelihood is
$$L(\lambda\mid \boldsymbol y,\boldsymbol\delta)=\prod_{i=1}^5 (\lambda e^{-\lambda y_i})^{\delta_i}(e^{-15\lambda})^{1-\delta_i} \tag{2}$$
Note that $(1)$ and $(2)$ are equivalent.
From $(1)$, it follows from usual calculus that the ML estimate of $\lambda$ is
$$\hat\lambda(\boldsymbol x)=\frac{4}{\sum_{i=1}^4 x_i+15}$$
