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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
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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
source | link

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: $$f(x)=\int_0^\infty f(x\mid\theta)\pi(\theta)\,d\theta = \int_x^\infty \frac{1}{\theta}\pi(\theta)\,d\theta\, .$$