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mathmari

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Hey!

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

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

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

- 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? $$$$ - $P(A\mid B)=1-P(A\mid \overline{B})$, 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?

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