# conditional probability of continuous independent random variables

I would like to know what's the exact way to obtain the conditional probabilities of node having multiple independent continuous random variables as its parent.

Say something like a Noisy OR gate model, with multiple parents of continuous random variable types.

The samples, I stumbled upon so far (for Noisy OR gate) where explained using discrete binary random variables, but my interest mostly lies on achieving similar outcomes with one or more continuous & random as inputs?

Let me explain with a simple example from Google results. A simple study says Bronchitis (B), Tuberculosis (T) and Lung cancer (L) causes Fatigue (F). It also says probability of B -> F = 0.6, T -> F = 0.7, L -> F = 0.8. with this information if we had to find out P(F|B,T,L), then the way to do this is find out

P(no F|B, T, L) = (1-0.6)(1-0.7)(1-0.8) = 0.024
then P(F|B,T,L) = 1 - P(no F|B, T, L) = 0.976.

In the above case all the causation random variable are of type discrete. In case if we add a continuous random variable say age to the scenario.

B -> F = 0.6, T -> F = 0.7, L -> F = 0.8
and if age (a) is
a <= 30 -> F = 0.1,
30 > a <= 50 -> F = 0.2, and
a > 50 -> F = 0.7.

Then if I had to find out P(F|B,T,L,30 > a <= 50), then what should be logic to find it out?

If just look at the example without considering age, we never know the joint probability distribution of of two or more individual events, but the final conditional probability of F given B,L & C has happened at ones is obtained using the individual probabilities of B,L & C.

Similarly I would like to know the way to compute P(F|B,T,L,30 > a <= 50)? Where 'a' is a discretized continuous random variable to ease the analysis.

Thanks, Kamal.

• In the title of the post it looks like you're asking for the joint probability, but in the first sentence you've written that you're looking for conditional probabilities, so I may be misunderstanding your question, but both are pretty easy to compute for independent continuous random variables. $f(x,y)=f_X(x)f_Y(y)$ and $f_{X|Y}(X|Y)=f_X(x)$. May 21 '14 at 18:00
• Thanks for pointing out the mistake, I was looking for conditional probability. May I request you to help me with an example. May 22 '14 at 5:22
• Thanks for the edits, they've made things clearer. The event that $30\lt a\le 50$ can be treated just like B, T, and L. $P(F|B,T,L,30\lt a\le 50) = 1-(1-0.6)(1-0.7)(1-0.8)(1-0.2)=0.9808$. May 22 '14 at 6:14
• Hi, This is where I was having my doubt, can I just pick (1−0.2) as such! Let me modify the example a bit for my clarification. Say we have 100 data points measuring F to Yes or No just because of age. Upon grouping them by age, we see the 1st 29 are for age <30 and it says F=Y for 16 and F=N for 13 data points. 2nd 30 (for 30 >age< 50) says F=Y for 8 and F=N for 22. The last 41 (for age > 50) says F=Y for 17 and F=N for 24. (continued)=> May 22 '14 at 9:14
• If we draw a cross table of probabilities, we see probabilities of F=Y & N across ages as 0.16, 0.13 | 0.08, 0.22 | 0.17, 0.24. Now assume one fine day 3 patients across different age groups got admitted and if a Nurse has to answer the probability of F? then as per our computation F=> 1 - (1-0.16)*(1-0.08 )*(1-0.17) = 0.359, which is less than sum of all the individual probabilities 0.16 + 0.08 + 0.17 = 0.41. Am not sure which is correct either 0.359 or sum of the independent individual probabilities 0.41. May 22 '14 at 9:15

Let's denote the presence of a cause of fatigue by some letter $X$ and the event that that cause leads to fatigue by $X_f$, so that the probability that a cause of fatigue leads to fatigue is $P(X_f)$. Given the scenario you describe regarding the independent causes of fatigue, $F$, then you're right to say $$P(F|B,T,L)=1-(1-P(B_f))(1-P(T_f))(1-P(L_f))$$
Now we consider an additional cause to fatigue that is independent of the other causes, in this case age $A$, that has different probabilities of leading to fatigue depending it's value. Given that $P(A_f|A\lt 30)=0.1, P(A_f|30\lt A\le 50)=0.2$, and $P(A_f|A\gt 50)=0.7$, we can now treat each of these cases as causes themselves that are independent to $B, L,$ and $T$. So, for example $$P(F|B,T,L,30\lt A\le 50)=1-(1-P(B_f))(1-P(T_f))(1-P(L_f))\left(1-P(A_f|30\lt A\le 50)\right)$$
In the other scenario you describe in the comments above we have the following data (modified slightly so that it adds up): 16 of the first 29 ages lead to $F$, 8 of the next 21 ages lead to $F$, and 17 of the last 50 ages lead to $F$. If we assume that each patient is equally likely to be any of the ages within their age group then we can write that data like this: $$P(A_f|A\lt 30)=\frac{16}{29}$$$$P(A_f|30\le A\le 50)=\frac{8}{21}$$$$P(A_f|A\gt 50)=\frac{17}{50}$$ Now this is the same as the original scenario in your question and we'd still be able to use the same general formula, for example: $$P(F|B,T,L,30\le A\le 50)=1-\left(1-P(B_f)\right)\left(1-P(T_f)\right)\left(1-P(L_f)\right)\left(1-P(A_f|30\le A\le 50)\right)$$
If three patients, one in each age group, came to see a nurse then the probability that one or more of them had fatigue based solely on their age would be: $$1-\left(1-P(A_f|A\lt 30)\right)\left(1-P(A_f|30\le A\le 50)\right)\left(1-P(A_f|A\gt 50)\right)=$$ $$1-\left(1-\frac{16}{29}\right)\left(1-\frac{8}{21}\right)\left(1-\frac{17}{50}\right)\approx0.817$$
Let's say that $A$ isn't a set of discrete points, though, and instead say that it's modeled by a continuous distribution. Then we'd have a probability density function, $f_A(a)$, such that $P(A_f|s\lt A\lt t)=\int_s^t{f_A(a)}da$ and we'd still be able to use the same general formula as we did in the discrete case: $$P(F|B,T,L,30\le A\le 50)=1-\left(1-P(B_f)\right)\left(1-P(T_f)\right)\left(1-P(L_f)\right)\left(1-\int_{30}^{50}f_A(a)da\right)$$