Showing that $X_{(1)}$ is sufficient for shifted exponential distribution

Question

If the pdf of a random sample is $f(x)=e^{-(x-θ)}$ where $x \geq θ$, Show that $T=X_{(1)}$ is a sufficient statistic for $θ$.

Can one show that $T$ is a sufficient statistic for $θ$ in the following manner?

Step 1: Find the pdf of $T$ $$f_{x_{(1)}}(x_{(1)})=ne^{-n(x_{(1)}-θ)} \quad x_{(1)} \geq θ$$

Step 2: Find the joint density of the random variables $$f_θ(x_1,\ldots,x_ n)=e^{-\sum_{i=1}^{n} (x_i-θ)} \quad x_i \geq θ $$

Step 3: Factor the joint density where we can extract the statistic in question $$f_θ(x_1,\ldots,x_n)=e^{-(x_{(1)}-nθ)}\cdot e^{-\sum_{i=2}^n x_{(i)}}$$

Step 4: Manipulate the above expression so that one factor is the pdf of the sufficient statistic and the other a factor not depending on $θ$

$$f_θ(x_1,\ldots,x_n)=ne^{-n(x_{(1)}-θ)} \cdot \frac{1}{n} e^{-\sum_{i=2}^{n} (x_{(i)}-x_{(1)})}$$

Step 5: By the Neyman-Fisher factorization criterion $X_{(1)}$ is sufficient for $θ$.

The above solution was given in my class. However, I'm not so sure it constitutes a proper proof/explanation as it does not make use of the indicator function in the joint pdf. Does the fact that we can manipulate the joint pdf into a product containing the pdf of the sufficient statistic somehow circumvents the need to use the indicator function?

Answer 1

$\require{cancel}$ $$ \xcancel{\vphantom{\sum^\sum_\sum} f_θ(x_1,\ldots,x_n) = e^{-\sum_{i=1}^n (x_i-θ)} \quad x_i \geq \theta} $$ Instead of writing the joint density like that, do it like this: $$ f_θ(x_1,\ldots,x_n) = e^{-\sum_{i=1}^n (x_i-θ)} \quad \theta \le x_{(1).} $$ (And notice that you incorrectly used capital letters on the left side.)

That shows the way in which it depends on $x_{(1)}.$

Then you have $$ f_\theta(x_1,\ldots,x_n) = \underbrace{ \vphantom{\mathbf 1(\theta\le x_{(1)})} e^{-\sum_{i=1}^n x_i} } \,\, \underbrace{ e^{n\theta} \cdot \mathbf 1(\theta\le x_{(1)})}. $$ where $$ \mathbf 1(\theta\le x_{(1)}) = \begin{cases} 1 & \text{if } \theta\le x_{(1)}, \\ 0 & \text{otherwise.} \end{cases} $$ Then you have one factor that depends on $(x_1,\ldots,x_n)$ only through the minimum $x_{(1)}$ and another that does not depend on $\theta.$

Answer 2

Does the fact that we can manipulate the joint pdf into a product containing the pdf of the sufficient statistic somehow circumvents the need to use the indicator function?

It is no circumvention. You have indirectly these indicator functions as a condition for the function (highlighted in red below).

$$f_{x_{(1)}}(x_{(1)})=ne^{-n(x_{(1)}-θ)} \quad {\color{red} {x_{(1)} \geq θ}}$$

which you can read as short-handed for

$$f_{x_{(1)}}(x_{(1)})= \begin{cases} ne^{-n(x_{(1)}-θ)} & \quad \text{if } { {x_{(1)} \geq θ}} \\ 0 & \quad \text{if } { {x_{(1)} < θ}} \end{cases}$$

and can be written with an indicator function as

$$f_{x_{(1)}}(x_{(1)})= ne^{-n(x_{(1)}-θ)} \cdot \mathbf{1}_{x_{(1)} \geq θ} $$

Manipulate the above expression so that one factor is the pdf of the sufficient statistic and the other a factor not depending on θ

Note that the factorisation does not neccesarily need to result in a factor that is the pdf of the sufficient statistic. It can indeed always be written like that but it is not necessary. Having a look at the pdf of the potential sufficient statistic might help, but the factorisation can sometimes be easier when we do not force it to be in that particular way.

In your case the formula

$$f_θ(x_1,\ldots,x_n)=ne^{-n(x_{(1)}-θ)} \cdot \frac{1}{n} e^{-\sum_{i=2}^{n} (x_{(i)}-x_{(1)})}$$

is equal to

$$f_θ(x_1,\ldots,x_n)=e^{nθ} \cdot e^{-\sum_{i=1}^{n} x_{(i)}}$$

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As a side note, intuitively we can reason that this is the distribution of an exponential distributed variable with known $\lambda = 1$ that is truncated at $\theta$.

In general, any distribution (with fixed parameters) that is truncated like that has $X_{(1)}$ as sufficient statistic. Other than the minimum value, there isn't any information in the sample about the truncation point.

Consider for example the following way to generate the data.

Compute uniform variables (representing quantiles)

$$Q_i \sim \mathcal{U}(q_t,1)$$

where $q_t = F(\theta)$ is the quantile where the truncation occurs.

And use the inverse transform to obtain $X_i$

$$X_i := F^{-1}(Q_i)$$

The latter transformation, defining the final distribution of the sample points, is irrelevant for the truncation point. And so the distribution of the sample has no information about the truncation point.

Related is a description of the German tank problem here. Describing why

why the samples {1, 2, 10} and samples {8, 9, 10} have the same solution