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I've been looking at this proof for a while, and I think I was able to answer the first part of the question. It asks:

(i) For any events $X$ and $Y$ with $\textbf{P}(Y) > 0$, show that $\textbf{P}(X|Y) + P(X^{c}|Y) = 1$.

My proof is as follows:

  • $\textbf{P}(X|Y) + \textbf{P}(X^{c}|Y ) = \textbf{P}(A\cap B)/\textbf{P}(B) + \textbf{P}(A^{c}\cap B)/\textbf{P}(B)$
  • $\textbf{P}(A\cap B) + \textbf{P}(A^{c}\cap B) = \textbf{P}(B)$
  • $\textbf{P}(A\cap B)/\textbf{P}(B) + \textbf{P}(A^{c} \cap B)/\textbf{P}(B) = (\textbf{P}(A\cap B) + \textbf{P}(A^{c} \cap B))/\textbf{P}(B)$
  • So, $(\textbf{P}(A \cap B) + \textbf{P}(A^{c} \cap B))/\textbf{P}(B) = \textbf{P}(B)/\textbf{P}(B) = 1$

(ii) If $\textbf{P}(Y|X) > P(Y)$, show that $\textbf{P}(Y^{c}|X) < P(Y^{c})$ by using (i).

I have no idea how to prove this, though. Any help (on either part) would be greatly appreciated.

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  • $\begingroup$ Have you tried proof by contradiction, say? $\endgroup$ – saulspatz Feb 14 at 0:55
  • $\begingroup$ @saulspatz No, I haven't. To be honest, I wouldn't really know a great place to start with that, either. $\endgroup$ – TheFiveHundredYears Feb 14 at 1:02
  • $\begingroup$ Why did you jump from using $X$ and $Y$ to using $A$ and $B$? Do not do that. $\endgroup$ – Graham Kemp Feb 14 at 1:07
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For the second part, switch $X$ and $Y$. By assumption,

$$ \mathbb{P}(Y|X) > \mathbb{P}(Y) $$ which, by (i) $$ \Rightarrow 1 - \mathbb{P}(Y'|X) > \mathbb{P}(Y) \Rightarrow \mathbb{P}(Y'|X) < 1 - \mathbb{P}(Y) = \mathbb{P}(Y'). $$

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  • $\begingroup$ Makes perfect sense, thank you! $\endgroup$ – TheFiveHundredYears Feb 14 at 1:21
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Given $\def\P{\mathsf P} \P(Y)>0$ then your proof is:

$$\begin{align}\P(X\mid Y)+\P(X'\mid Y)&=\dfrac{\P(X\cap Y)}{\P(Y)}+\dfrac{\P(X'\cap Y)}{\P(Y)}&&\text{definition of conditioning}\\&=\dfrac{\P((X\cap Y)\cup(X'\cap Y))}{\P(Y)}&&\text{additivity over disjoint events}\\&=\dfrac{\P((X\cup X')\cap Y)}{\P(Y)}&&\text{distribution}\\&=\dfrac{\P(Y)}{\P(Y)}&&\text{union of complements}\\&=1 &&\text{since }\P(Y)>0\end{align}$$

Similarly you can show the following: $$\begin{align}\P(Y\mid X)+\P(Y'\mid X)&=1\\\P(Y)+\P(Y')&=1\end{align}$$

With a little rearrangement: $$\begin{align}\P(Y\mid X)&=1-\P(Y'\mid X)\\\P(Y)&=1-\P(Y')&\end{align}$$


Now, when $\P(Y\mid X)>\P(Y)$, then ...

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