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The Question
I have a sample X1,...,Xn i.i.d. drawn from a uniform distribution $unif[0,\theta]$, θ ∈ Θ = R+;

And I'd just like to compute the joint PDF


The Solution
I have the following solution supplied to me: $$ f(\mathbf{X}:\theta) = \left( \frac{1}{\theta} \right)^{n} 1_{\{ X_{(n)} \leq \theta \}} $$

Where $X_{(n)} = max_{i=1,2,\ldots,n} \{ X_{1}, \ldots, X_{n} \}$


Where I'm Stuck
My PDF for this distribution will be: $$ f(X) = \left\{ \begin{matrix} \frac{1}{\theta}&, X_{i} \leq \theta \\ 0&, X_{i} > \theta \end{matrix}\right. $$

As the observations are independent, my joint distribution is simply: $$ \begin{align*} l(X:\theta) &= \prod^{n}_{i=1} f(X_{i})\\ &= \prod^{n}_{i=1} \frac{1}{\theta} \cdot 1_{\{X_{i} \leq \theta\}} \\ &= \left( \frac{1}{\theta} \right) ^{n} \prod^{n}_{i=1} 1_{\{X_{i} \leq \theta\}} \\ \end{align*} $$

Is the supplied solution just saying that we need all observations to be less than $\theta$ for the joint PDF to yeild something non-trivial?

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  • 2
    $\begingroup$ The supplied solution is incorrect. The indicator also needs to specify that all $X_i$ are non-negative. Equivalently, it would suffice to stipulate $X_{(1)}\ge 0$. Your answer $\prod_{i=1}^n f(X_i)$ is correct but then the same mistake creeps into your use of the indicator function. (Note that $f(X) = (1/\theta)I(0\lt X \le \theta)$.) $\endgroup$ – whuber Apr 13 '15 at 17:42
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A product of indicator functions equals 1 if and only if every indicator in the product equals 1. In this case your product of indicators equals 1 iff all $X_i \leq \theta$ which is equivalent to the maximum, $X_{(n)}$, being less than or equal to $\theta$.

What you have is identical to the answer, just in a less compact form. Beyond simple convenience, we also like to write it in the way that the solution gives because that points to the fact that $X_{(n)}$ is sufficient for $\theta$.

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