0
$\begingroup$

If $A$ is a subset of $R$ and $X$ is a random variable. I have two variables $X_1$ and $X_2$. $I$ being $1$ if $X$ in subset $A$ and $0$ if not in $A$. Let $U$~$U(0;1)$ and determine if this pair is independent. Verify your claim using simulation in Matlab.

$$ X_1 = I_U \epsilon\left[\left.0,\frac{1}{3}\right.\right), X_2 = I_U\epsilon\left[\left.\frac{1}{3},\frac{2}{3}\right.\right)$$

I usually show my work done, but I cannot find how to determine if these are independent. My question: Please, can someone explain how to show if this pair is independent? From there, then I can attempt how to verify using Matlab.

$\endgroup$
  • 1
    $\begingroup$ You might start by carefully writing down the definition of independence: what does it mean for two random variables to be independent? $\endgroup$ – Xander Henderson Apr 21 at 1:29
1
$\begingroup$

Intuitively, if we know that $X_1$ is one, then we know that $X_2$ is zero, so knowing $X_1$ means we know something about $X_2$, which suggests that they are not independent.

Explictly:

Let $A=\{1\}$. Compute $P[X_1 \in A], P[X_2 \in A], P[X_1 \in A, X_2 \in A]$.

Is $P[X_1 \in A, X_2 \in A] = P[X_1 \in A] P[X_2 \in A]$?

$\endgroup$
  • $\begingroup$ I'm sorry. I am just getting caught up on notation and the simplest things. I am thrown off by the U being uniform (0,1), and then $X_1$ and $X_2$ have the additional U notation subscript. Is this correct: $f(x_1) =$ integral from 0 to 1/3 $(f(x)dx)$, but here $f(x)$ is just 1, so $f(x_1) = 1/3$. Same for $f(x_2)$, so $f(x_2) = 2/3-1/3 = 1/3$. $1/3*1/3 =1/9$, which is $P(x_1,x_2) $and it does equal to$ P(x_1)*P(x_2), 1/3*1/3 = 1/9$. Or did I just compute the same thing 2 different ways? $\endgroup$ – PattyWatty27 Apr 21 at 4:40
  • 1
    $\begingroup$ The notation is a little confusing. The variable $U$ is uniformly distributed on $[0,1]$. $X_1$ is one if $U \in [0,{1 \over 3})$ and zero otherwise. Similarly for $X_2$. You can just compute the probabilities directly, $P[X_1 =1] = P[U \in [0,{1 \over 3})] = {1 \over 3}$ and similarly for $X_2$. However, $P[X_1=1, X_2 = 1] = 0$. $\endgroup$ – copper.hat Apr 21 at 4:58
  • $\begingroup$ Ok, so if I changed $X_2$ to $[\frac{2}{9}, \frac{2}{3})$: $P[X_1 =1]=P[U∈[0,\frac{1}{3} )]=\frac{1}{3}$ and $P[X_2 =1]=P[U∈[\frac{2}{9},\frac{5}{9} )]=\frac{1}{3}$, and the $P[X_1 =1,X_2 =1]=\frac{1}{9}$. which equals $P(X_1)*P(X_2) = \frac{1}{3}*\frac{1}{3} = \frac{1}{9}$, therefore, they are independent. And say I changed $X_2$ again, $P[X_2 =1]=P[U∈[\frac{2}{9},\frac{2}{3} )]=\frac{4}{9}$. In this case, it still holds that $P[X_1 =1,X_2 =1]=\frac{1}{9}$, but this does not equal $P(X_1)*P(X_2) = \frac{1}{3}*\frac{4}{9} = \frac{4}{27}$, therefore these would be dependent? $\endgroup$ – PattyWatty27 Apr 21 at 5:23
  • $\begingroup$ Sorry, you lost me there. What are you trying to do? For $X_1,X_2$ to be independent you must show that $P[X_1 \in A, X_2 \in B] = P[X_1 \in A] P[X_2 \in B]$ for all $A,B$. $\endgroup$ – copper.hat Apr 21 at 5:40
  • $\begingroup$ Sorry! I just wanted to make sure I understood you correctly, so I created 2 new scenarios. I just changed the bounds on $X_2$ given initially and recomputed the problem. $\endgroup$ – PattyWatty27 Apr 21 at 5:44

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.