I am studying for my machine learning exam and i have problems with basic conditional probability. I have to solve the following exercise:

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The formular for conditionally independence is

$P(Y|x) = \prod_{i=1}^{m} P(Y_{i}|x)$ and also

$P(Y|x) = P(Y_{1}|x) \prod_{i=1}^{m} P(Y_{i}|Y_{1},...,Y_{i-1},x)$

So i thought about the problem and came to these conclusions:

$P(Y_{1}|x) = 0.6$ because given an $x$ the probability of $Y_{1}$ being 1 is $0.3$ plus $0.3$. I've done the same for

$P(Y_{2}|x) = 0.4$

$P(Y_{3}|x) = 0.3$

So now i can compute $\prod_{i=1}^{m} P(Y_{i}|x)$, but what is $P(Y|x)$?

I thought that i can get $P(Y|x)$ by the second formula, but for that i require $P(Y_{1}|x)$, $P(Y_{2}|Y_{1},x)$, $P(Y_{3}|Y_{2},Y_{1},x)$.

For $P(Y_{2}|Y_{1},x)$ i came up with this:

$P(Y_{2}|Y_{1},x) = \frac{P(Y_{2}|x)}{P(Y_{1})}$, and $P(Y_{1}) = P(Y_{1}|x) P(X)$

But what is $P(X)$? How can i solve this problem? Are there any mistakes?

  • $\begingroup$ Please add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. $\endgroup$ – gung - Reinstate Monica Aug 8 '18 at 14:43

You already computed:







The values of $P(Y|x)$ that you are asking about (i.e. the joint probabilities) are actually given to you in the last column of the table as $P(y_1,y_2,y_3|x)$. So the final step is to check if the products are equal to the values of that column.

For example, row 4 of the table says that for $\{y_1=1,y_2=0,y_3=0\}$ we should get a product of $0.3$ if there is indeed conditional independence. So let's check:

$P(y_1=1|x) \times P(y_2=0|x) \times P(y_3=0|x)=0.6 \times 0.6 \times 0.7 = 0.252 \neq 0.3$

It appears that $P(y_1|x)P(y_2|x)P(y_3|x) \neq P(y_1,y_2,y_3|x)$

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  • $\begingroup$ Ah. That makes sense. I was already wondering why i had $P(Y|x)$, but i indeed calculated it for $P(Y_{1}= 1|x)$ $\endgroup$ – Jens Aug 8 '18 at 14:52

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