0
$\begingroup$

Let there be two random variables $X_1$ and $X_2$. With probability $\gamma$, they are perfectly correlated and each distributed uniformly on the interval $[0,1]$. With probability $1-\gamma$, they are independent and each distributed uniformly on the interval $[0,1]$.

  1. What is the joint density function? i.e. $f(x_1,x_2)$?

  2. What is the distribution function? i.e. $Prob(X_1 \leq x_1, X_2 \leq x_2):=F(x_1,x_2)$?

  3. What is the conditional distribution function? i.e. $Prob(X_1 \leq x_1|x_2):=F(x_1|x_2)$?

For 1), I tried something like \begin{align} f(x,y)=\begin{cases} 1 &\text{ if } x=y \\ 1-\gamma &\text{ if } x\neq y \end{cases} \end{align}

However, it does not seems right. Because when I do the integration (let $A=\{(a,b)\in [0,1]^2:a=b \}$)

\begin{align} &\int\int_{(x,y)\in [0,1]^2}f(x,y)dxdy\\ =& \int\int_{(x,y)\in [0,1]^2 \setminus A}f(x,y)dxdy + \int\int_{(x,y)\in A }f(x,y)dxdy \\ =& \int\int_{(x,y)\in [0,1]^2 \setminus A}(1- \gamma) dxdy + \int\int_{(x,y)\in A }1 dxdy \\ \end{align}

where $\int\int_{(x,y)\in [0,1]^2 \setminus A} (1- \gamma) dxdy =(1- \gamma)$ ? and $\int\int_{(x,y)\in A }1 dxdy =0$?

If it is true, it does not add up to 1... I don't know where it goes wrong....

Thank you in advance!!!

$\endgroup$
8
  • $\begingroup$ Welcome to Mathematics SE. Take a tour: math.stackexchange.com/tour. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context by stating what you understand about the problem, what you've tried so far, etc.; both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance: math.meta.stackexchange.com/q/9959 . $\endgroup$
    – Kavi Rama Murthy
    Commented Mar 10, 2022 at 4:49
  • $\begingroup$ Hint: Use the Law of Total Probability. Hint 2: Consider the cases $x_1=x_2$ and $x_1\neq x_2$. $\endgroup$
    – Graham Kemp
    Commented Mar 10, 2022 at 4:57
  • $\begingroup$ Hi Graham, thanks for your help! I actually tried... I edited the question. I still have trouble.... $\endgroup$
    – Paul H.Y. Cheung
    Commented Mar 10, 2022 at 5:21
  • 1
    $\begingroup$ Thank you very much for the hint and following-up! I appreciate it. I am now convinced that the distribution does not have a pdf. For conditional distribution function, I am thinking $X_2=x_2$. I think it might look like \begin{align*} F(x_1|x_2)=\begin{cases} \gamma + (1-\gamma)x_1 & \text{ if } x_1 \geq x_2 \\ (1-\gamma)x_1 & \text{ if } x_1 < x_2 \end{cases} \end{align*} For the distribution function, is it something like $$F(x_1,x_2)=\gamma \min\{x_1,x_2\} + (1-\gamma)x_1x_2 ?$$ $\endgroup$
    – Paul H.Y. Cheung
    Commented Mar 11, 2022 at 16:35
  • 1
    $\begingroup$ The idea of $\min$ is kind of coming from your hint. I just check the wikipedia page about Dirac Delta function. It looks like it is related to the $\textbf{Hint:}$ that you wrote! I think you are probably saying there is a way to approximate the pdf using the limit of the some function such that the limit of that function will be almost everywhere close to the pdf I look for? $\endgroup$
    – Paul H.Y. Cheung
    Commented Mar 11, 2022 at 16:38

0

Reset to default

You must log in to answer this question.