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 3 deleted 10 characters in body edited Sep 10 '12 at 1:46 Zen 17.8k33 gold badges5757 silver badges100100 bronze badges What you're missing isYou seem to be a proper understanding oflittle confused about the likelihood function of the $$U[0,\theta]$$ model. Let $$f(x\mid\theta)=1/\theta$$, for $$0\leq x\leq\theta$$, and $$f(x\mid\theta)=0$$, otherwise, for $$\theta>0$$. For some fixed $$x$$, what is the graph of $$f(x\mid\theta)$$ as a function of $$\theta$$? To draw the graph, notice -- and this is the key point -- that $$x\in[0,\theta]$$ if and only if $$\theta\in[x,\infty)$$. So, using indicator functions, we have $$f(x\mid\theta)=\frac{1}{\theta}I_{[0,\theta]}(x)=\frac{1}{\theta}I_{[x,\infty)}(\theta)\, .$$ After you understand this, just do the integration: $$f(x)=\int_0^\infty f(x\mid\theta)\pi(\theta)\,d\theta = \int_x^\infty \frac{1}{\theta}\pi(\theta)\,d\theta\, .$$ What you're missing is a proper understanding of the likelihood function of the $$U[0,\theta]$$ model. Let $$f(x\mid\theta)=1/\theta$$, for $$0\leq x\leq\theta$$, and $$f(x\mid\theta)=0$$, otherwise, for $$\theta>0$$. For some fixed $$x$$, what is the graph of $$f(x\mid\theta)$$ as a function of $$\theta$$? To draw the graph, notice -- and this is the key point -- that $$x\in[0,\theta]$$ if and only if $$\theta\in[x,\infty)$$. So, using indicator functions, we have $$f(x\mid\theta)=\frac{1}{\theta}I_{[0,\theta]}(x)=\frac{1}{\theta}I_{[x,\infty)}(\theta)\, .$$ After you understand this, just do the integration: $$f(x)=\int_0^\infty f(x\mid\theta)\pi(\theta)\,d\theta = \int_x^\infty \frac{1}{\theta}\pi(\theta)\,d\theta\, .$$ You seem to be a little confused about the likelihood function of the $$U[0,\theta]$$ model. Let $$f(x\mid\theta)=1/\theta$$, for $$0\leq x\leq\theta$$, and $$f(x\mid\theta)=0$$, otherwise, for $$\theta>0$$. For some fixed $$x$$, what is the graph of $$f(x\mid\theta)$$ as a function of $$\theta$$? To draw the graph, notice -- and this is the key point -- that $$x\in[0,\theta]$$ if and only if $$\theta\in[x,\infty)$$. So, using indicator functions, we have $$f(x\mid\theta)=\frac{1}{\theta}I_{[0,\theta]}(x)=\frac{1}{\theta}I_{[x,\infty)}(\theta)\, .$$ After you understand this, just do the integration: $$f(x)=\int_0^\infty f(x\mid\theta)\pi(\theta)\,d\theta = \int_x^\infty \frac{1}{\theta}\pi(\theta)\,d\theta\, .$$ 2 edited body edited Feb 27 '12 at 2:26 Zen 17.8k33 gold badges5757 silver badges100100 bronze badges What you're missing is a proper understanding of the likelihood function of the $$U[0,\theta]$$ model. Let $$f(x\mid\theta)=1/\theta$$, for $$0\leq x\leq\theta$$, and $$f(x\mid\theta)=0$$, otherwise, for $$\theta>0$$. For some fixed $$x$$, what is the graph of $$f(x\mid\theta)$$ as a function of $$\theta$$? To draw the graph, notice -- and this is the key point -- that $$x\in[0,\theta]$$ if and only if $$\theta\in[x,\infty]$$$$\theta\in[x,\infty)$$. So, using indicator functions, we have $$f(x\mid\theta)=\frac{1}{\theta}I_{[0,\theta]}(x)=\frac{1}{\theta}I_{[x,\infty]}(\theta)\, .$$$$f(x\mid\theta)=\frac{1}{\theta}I_{[0,\theta]}(x)=\frac{1}{\theta}I_{[x,\infty)}(\theta)\, .$$ After you understand this, just do the integration: $$f(x)=\int_0^\infty f(x\mid\theta)\pi(\theta)\,d\theta = \int_x^\infty \frac{1}{\theta}\pi(\theta)\,d\theta\, .$$ What you're missing is a proper understanding of the likelihood function of the $$U[0,\theta]$$ model. Let $$f(x\mid\theta)=1/\theta$$, for $$0\leq x\leq\theta$$, and $$f(x\mid\theta)=0$$, otherwise, for $$\theta>0$$. For some fixed $$x$$, what is the graph of $$f(x\mid\theta)$$ as a function of $$\theta$$? To draw the graph, notice -- and this is the key point -- that $$x\in[0,\theta]$$ if and only if $$\theta\in[x,\infty]$$. So, using indicator functions, we have $$f(x\mid\theta)=\frac{1}{\theta}I_{[0,\theta]}(x)=\frac{1}{\theta}I_{[x,\infty]}(\theta)\, .$$ After you understand this, just do the integration: $$f(x)=\int_0^\infty f(x\mid\theta)\pi(\theta)\,d\theta = \int_x^\infty \frac{1}{\theta}\pi(\theta)\,d\theta\, .$$ What you're missing is a proper understanding of the likelihood function of the $$U[0,\theta]$$ model. Let $$f(x\mid\theta)=1/\theta$$, for $$0\leq x\leq\theta$$, and $$f(x\mid\theta)=0$$, otherwise, for $$\theta>0$$. For some fixed $$x$$, what is the graph of $$f(x\mid\theta)$$ as a function of $$\theta$$? To draw the graph, notice -- and this is the key point -- that $$x\in[0,\theta]$$ if and only if $$\theta\in[x,\infty)$$. So, using indicator functions, we have $$f(x\mid\theta)=\frac{1}{\theta}I_{[0,\theta]}(x)=\frac{1}{\theta}I_{[x,\infty)}(\theta)\, .$$ After you understand this, just do the integration: $$f(x)=\int_0^\infty f(x\mid\theta)\pi(\theta)\,d\theta = \int_x^\infty \frac{1}{\theta}\pi(\theta)\,d\theta\, .$$ 1 answered Feb 27 '12 at 2:16 Zen 17.8k33 gold badges5757 silver badges100100 bronze badges What you're missing is a proper understanding of the likelihood function of the $$U[0,\theta]$$ model. Let $$f(x\mid\theta)=1/\theta$$, for $$0\leq x\leq\theta$$, and $$f(x\mid\theta)=0$$, otherwise, for $$\theta>0$$. For some fixed $$x$$, what is the graph of $$f(x\mid\theta)$$ as a function of $$\theta$$? To draw the graph, notice -- and this is the key point -- that $$x\in[0,\theta]$$ if and only if $$\theta\in[x,\infty]$$. So, using indicator functions, we have $$f(x\mid\theta)=\frac{1}{\theta}I_{[0,\theta]}(x)=\frac{1}{\theta}I_{[x,\infty]}(\theta)\, .$$ After you understand this, just do the integration: $$f(x)=\int_0^\infty f(x\mid\theta)\pi(\theta)\,d\theta = \int_x^\infty \frac{1}{\theta}\pi(\theta)\,d\theta\, .$$