MHB Conditional probability prove or disprove

[mathmari]

Hey!

Let $P$ be a probability measure on a $\sigma$-Algebra $\mathcal{A}$. I want to prove or disprove the following statements:

1. $P(A\mid B)=1-P(\overline{A}\mid B)$, for $A, B\in \mathcal{A}$
2. $P(A\mid B)=1-P(A\mid \overline{B})$, for $A, B\in \mathcal{A}$

I have done the following:

1. We have that $B=\left (A\cap B\right )\cup \left (\overline{A}\cap B\right )$. Since the sets $\left (A\cap B\right )$ and $\left (\overline{A}\cap B\right )$ are disjoint we have that $P\left [\left (A\cap B\right )\cup \left (\overline{A}\cap B\right )\right ]=P\left (A\cap B\right )+P \left (\overline{A}\cap B\right )$.
   Therefore, we get $P(B)=P\left (A\cap B\right )+P \left (\overline{A}\cap B\right )$.
   
   We have that $P(A\mid B)=\frac{P(A\cap B)}{P(B)}\Rightarrow P(A\cap B)=P(A\mid B)P(B)$ and $P(\overline{A}\mid B)=\frac{P(\overline{A}\cap B)}{P(B)}\Rightarrow P(\overline{A}\cap B)=P(\overline{A}\mid B)P(B)$.
   
   So, we get $P(B)=P(A\mid B)P(B)+P(\overline{A}\mid B)P(B) \Rightarrow P(B)=P(B)\left [P(A\mid B)+P(\overline{A}\mid B)\right ] \overset{P(B)>0}{\Rightarrow }1=P(A\mid B)+P(\overline{A}\mid B) \Rightarrow P(A\mid B)=1-P(\overline{A}\mid B)$
   So, the statement is true.
   
   Is this correct? $$$$
2. $P(A\mid B)=1-P(A\mid \overline{B})$, for $A, B\in \mathcal{A}$
   
   From the definition of conditional probability then we get the statement $\frac{P(A\cap B)}{P(B)}=1-\frac{P(A\cap \overline{B})}{P(\overline{B})}$.
   
   For $A=\Omega$ we get $\frac{P(\Omega\cap B)}{P(B)}=1-\frac{P(\Omega\cap \overline{B})}{P(\overline{B})}\Rightarrow \frac{P( B)}{P(B)}=1-\frac{P( \overline{B})}{P(\overline{B})}\Rightarrow 1=1-1 \Rightarrow 1=0$, a contradiction.
   
   Therefore, this statement is in general not true.
   
   Is this correct?

 

[I like Serena]

1. $P(A\mid B)=1-P(\overline{A}\mid B)$, for $A, B\in \mathcal{A}$

Hey mathmari!

What if $P(B)=0$? (Wondering)

2. $P(A\mid B)=1-P(A\mid \overline{B})$, for $A, B\in \mathcal{A}$

From the definition of conditional probability then we get the statement $\frac{P(A\cap B)}{P(B)}=1-\frac{P(A\cap \overline{B})}{P(\overline{B})}$.

How do we get that? (Wondering)

 

[mathmari]

What if $P(B)=0$? (Wondering)

If $P(B)=0$ then the conditional probaibility $P(A\mid B)$ is not defined, is it? (Wondering)

How do we get that? (Wondering)

By definition we have that $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$ and $P(A\mid \overline{B})=\frac{P(A\cap \overline{B})}{P(\overline{B})} $, or not? Therefore we get that $P(A\mid B)=1-P(A\mid \overline{B})$ is equivalent to $1-\frac{P(A\cap \overline{B})}{P(\overline{B})} $. Is this wrong? (Wondering)

 

[I like Serena]

If $P(B)=0$ then the conditional probaibility $P(A\mid B)$ is not defined, is it? (Wondering)

Ah yes.
That's because $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$, which is indeed undefined if $P(B)=0$.

By definition we have that $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$ and $P(A\mid \overline{B})=\frac{P(A\cap \overline{B})}{P(\overline{B})} $, or not?

Yes.

Therefore we get that $P(A\mid B)=1-P(A\mid \overline{B})$ is equivalent to $1-\frac{P(A\cap \overline{B})}{P(\overline{B})} $. Is this wrong?

Well, I'm only aware of the complement rule $P(\overline U)=1-P(U)$, but it's not clear to me how this is applied.

 

[mathmari]

Well, I'm only aware of the complement rule $P(\overline U)=1-P(U)$, but it's not clear to me how this is applied.

From the definition of the conditional probability we have that $P(A\mid B)=\frac{\mathbb{P}(A\cap B)}{\mathbb{P}(B)}$ and $P(A\mid \overline{B})=\frac{\mathbb{P}(A\cap \overline{B})}{\mathbb{P}(\overline{B})}$.

Replacing these at $\mathbb{P}(A\mid B)=1-\mathbb{P}(A\mid \overline{B})$ we get $\frac{\mathbb{P}(A\cap B)}{\mathbb{P}(B)}=1-\frac{\mathbb{P}(A\cap \overline{B})}{\mathbb{P}(\overline{B})}$.

Isn't it correct? (Wondering)

 

[I like Serena]

From the definition of the conditional probability we have that $P(A\mid B)=\frac{\mathbb{P}(A\cap B)}{\mathbb{P}(B)}$ and $P(A\mid \overline{B})=\frac{\mathbb{P}(A\cap \overline{B})}{\mathbb{P}(\overline{B})}$.

Replacing these at $\mathbb{P}(A\mid B)=1-\mathbb{P}(A\mid \overline{B})$ we get $\frac{\mathbb{P}(A\cap B)}{\mathbb{P}(B)}=1-\frac{\mathbb{P}(A\cap \overline{B})}{\mathbb{P}(\overline{B})}$.

Isn't it correct? (Wondering)

Oh wait! You're doing a proof by contradiction!
I totally missed that.
Perhaps good to mention that. That is, 'suppose it is true, then...'. (Nerd)

 

[mathmari]

Oh wait! You're doing a proof by contradiction!
I totally missed that.
Perhaps good to mention that. That is, 'suppose it is true, then...'. (Nerd)

Ah ok! Is the counterexample that I used correct? (Wondering)

 

[I like Serena]

Ah ok! Is the counterexample that I used correct?

Yes. (Nod)

 

[mathmari]

Yes. (Nod)

Great! Thank you! (Yes)