MHB For which values are the variables independent?

AI Thread Summary
The discussion focuses on determining the values of variables $c$ and $d$ for which random variables $X$ and $Y$ are independent. The calculations show that $c$ must equal $\frac{1}{8}$ and $d$ must equal $\frac{1}{4}$, as these values satisfy the independence condition $P[X=m, Y=n] = P[X=m] \cdot P[Y=n]$. Participants agree that checking all combinations may not be necessary, as independence of one pair implies independence of others. The conversation emphasizes the importance of confirming independence through selective checks rather than exhaustive calculations. Ultimately, the approach to verifying independence is validated by the calculations presented.
mathmari
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Hey! :giggle:

We have the table of distribution of $X$, $Y$ and their joint random variable :
1638005697962.png


with $$(c,d)\in \left \{(c,d)\in \mathbb{R}^2\mid 0\leq c\leq \frac{1}{4}, \ 0\leq d\leq \frac{1}{2}, \ \frac{1}{4}\leq c+d\leq \frac{1}{2}\right \}$$

I want to calculate the values of $c$ and $d$ such that $X$ and $Y$ are independent.
So do we have to check each combination so that $P[X=m,\ Y=n]=P[X=m]\cdot P[Y=n]$ ?

\begin{align*}&P[X=-1, \ Y=0]=c \ \text{ and } \ P[X=-1]\cdot P[Y=0]=\frac{1}{4}\cdot \frac{1}{2}=\frac{1}{8} \text{ so } c=\frac{1}{8} \\ &P[X=0, \ Y=0]=d \ \text{ and } \ P[X=0]\cdot P[Y=0]=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4} \text{ so } d=\frac{1}{4}\end{align*}
We check now also the others if $X$ and $Y$ are indeed independent (or do we not have to? :unsure: )
\begin{align*}&P[X=-1, \ Y=1]=\frac{1}{4}-c=\frac{1}{4}-\frac{1}{8}=\frac{1}{8} \ \text{ and } \ P[X=-1]\cdot P[Y=1]=\frac{1}{4}\cdot \frac{1}{2}=\frac{1}{8} \text{ so correct} \\ &P[X=0, \ Y=0]=d=\frac{1}{4} \ \text{ and } \ P[X=0]\cdot P[Y=0]=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4} \text{ so correct} \\ &P[X=0, \ Y=1]=\frac{1}{2}-d=\frac{1}{2}-\frac{1}{4}=\frac{1}{4} \ \text{ and } \ P[X=0]\cdot P[Y=1]=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4} \text{ so correct}\\ &P[X=1, \ Y=0]=\frac{1}{2}-c-d=\frac{1}{2}-\frac{1}{8}-\frac{1}{4}=\frac{1}{8} \ \text{ and } \ P[X=1]\cdot P[Y=0]=\frac{1}{4}\cdot \frac{1}{2}=\frac{1}{8} \text{ so correct}\\ &P[X=1, \ Y=1]=c+d-\frac{1}{4}=\frac{1}{8}+\frac{1}{4}-\frac{1}{4}=\frac{1}{8} \ \text{ and } \ P[X=1]\cdot P[Y=1]=\frac{1}{4}\cdot \frac{1}{2}=\frac{1}{8} \text{ so correct} \end{align*}:unsure:
 
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Hey mathmari!

Seems correct to me. (Nod)

We might skip a couple of them, since we know that if $A$ and $B$ are independent, that $A$ and $B^c$ are also independent. 🤔
 
Klaas van Aarsen said:
We might skip a couple of them, since we know that if $A$ and $B$ are independent, that $A$ and $B^c$ are also independent. 🤔

You mean to check only with $Y=0$ and not with $Y=1$ ? :unsure:
 
mathmari said:
You mean to check only with $Y=0$ and not with $Y=1$ ?
Indeed. (Nod)
 
Similarly we can skip $X=1$ after checking $X=-1$ and $X=0$.
We can prove it if we want to. 🤔
 
I was reading documentation about the soundness and completeness of logic formal systems. Consider the following $$\vdash_S \phi$$ where ##S## is the proof-system making part the formal system and ##\phi## is a wff (well formed formula) of the formal language. Note the blank on left of the turnstile symbol ##\vdash_S##, as far as I can tell it actually represents the empty set. So what does it mean ? I guess it actually means ##\phi## is a theorem of the formal system, i.e. there is a...

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