I'm struggling to come up with a well reasoned argument for this problem.

Let $\tau_1$ be the posteriori probability and let $L(r^*)$ be the risk classifier.

For this scenario, assume:

$X\in\mathbb{X}=[0,1],Y\in\{ 0,1 \}$

$\pi_y=P(Y=y)=1/2$ for $y\in{0,1}$

Now assume $P(Y=1|x)=\tau_1\in[0,1]$ is unknown and the PDFs $f(x|Y=0)$ and $f(x|Y=1)$ are also unknown. In other words the only things that are known is that $f(\centerdot|Y=y):\mathbb{X}\longrightarrow\mathbb{R}$ is a density function on $\mathbb{X}=[0,1]$ for each $y\in\{0,1\}$. In other words $f(x|Y=y)\ge0$ for all $x\in\mathbb{X}$ and $\int_\mathbb{X}f(x|Y=y)dx=1$.

From this information I'd like to deduce the min and max values of $L(r^*)$ and provide conditions on $\tau_1$, $f(\centerdot|Y=0)$ and $f(\centerdot|Y=1)$ in order to yield those values.

  • $\begingroup$ Take my answer for the first part here: stats.stackexchange.com/a/549963/144600 and find the extrema points by derivation and equating to 0. $\endgroup$
    – Spätzle
    Oct 27 '21 at 8:22
  • 1
    $\begingroup$ @Spätzle: Please note that we want the OP to add self-study tag themselves, see stats.meta.stackexchange.com/questions/5611/… Better to ask a standard comment as I do below $\endgroup$ Oct 27 '21 at 11:44
  • $\begingroup$ Please add the self-study tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. Please make these changes as just posting your homework & hoping someone will do it for you is grounds for closing. $\endgroup$ Oct 27 '21 at 11:45
  • $\begingroup$ @kjetilbhalvorsen this is one question in a series of many by this OP, all with the same setting and an increasing level of abstraction. He/she have made it clear its a class material they didn't manage to understand. $\endgroup$
    – Spätzle
    Oct 27 '21 at 13:23

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