Show that for every $p$, $0\leq p\leq 1$, the function $f(x)$ = $p*sin(x) +(1-p)*cos(x)$, $0\leq x \leq \pi/2 $, and $f(x)=0$ otherwise, is a density function. Find its CDF and use it to find all the medians.

I was able to prove it a density function and also was able to get the CDF which is $(p +\sin(x) -p(\sin(x) + \cos(x))$ ; $0\leq x \leq \pi/2$ and $1$ for $ x\geq \pi/2$ and $zero$ elsewhere. (I hope I am correct)

I don't understand the last part. What do they mean by finding all the medians?

  • 3
    $\begingroup$ It means to derive a formula that is valid for any $p$, like a median$(p)$ function. You know that the median of Uniform$(a,b)$ is $(b-a)/2$. Do the same for your situation. $\endgroup$ – Dave Jan 11 at 23:56
  • $\begingroup$ You probably should add the self study tag. $\endgroup$ – Michael R. Chernick Jan 12 at 0:04
  • $\begingroup$ I still can't figure it out. If I put it equal to 0.5, I have an equation with two variable p and x. How can I solve this? $\endgroup$ – Leaderboard281923 Jan 12 at 0:10
  • $\begingroup$ Solve for $x$ to get the value that is the median. Write $x$ as a function of $p$. And, yes, self-study rag, please. $\endgroup$ – Dave Jan 12 at 0:24
  • 1
    $\begingroup$ Cross-posted at math.stackexchange.com/questions/3505774/…. $\endgroup$ – JimB Jan 12 at 3:27

First median is defined where $F(x)=0.5$. Right now you have a CDF that is defined in terms of $p$ and $x$ and it is possible to define the median in terms of $p$(I did not check if your CDF is correct or not since it is your task). You can think of $p$ as a parameter and thus no need to worry if you have a definition that have $p$ in it

| cite | improve this answer | |
  • $\begingroup$ I can't solve the equation when I am equating it to 0.5 I tried substitution $sin$$ x=t$ but still nothing. $\endgroup$ – Leaderboard281923 Jan 12 at 19:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.