Create a dataset with conditional probability, extension

Question

Following my previous question (How to create a dataset with conditional probability?), now I want to create a dataset in R containing the information about TWO symptoms, and 1 disease.

Symptoms $S_1$ and $S_2$ and independent and conditional independent, so $P(S_1|S_2)=P(S_1)$ and $P(S_2|S_1)=P(S_2)$.

Suppose we have the following data:

$P(D)=0.003$

$P(S_1)=0.005$

$P(S_2)=0.008$

$P(S_1|D)=0.3$

$P(S_2|D)=0.25$

Following the method that Henry used to solve my previous question, I think that we have to complete this table:

                                     Disease
                                     Yes       No

             |  Symptom1 Yes   |     a         b
Symptom2 Yes |                 |
             |           No    |     c         d
-------------|-----------------|------------------
             |  Symptom1 Yes   |     e         f
Symptom2 No  |                 | 
             |           No    |     g         h

What I'd like to calculate, from the dataset, is the value of $P(D|S_1 and S_2)$. Also I need to calculate the probability of each event, in order to run some simulations.

I tried to find the value for a...h, but I didn't find enough relationship to solve the 8-parameter system.

What I know is:

$P(D)=0.003 \rightarrow$ a + c + e + g = 0.003

$P(S_1)=0.005 \rightarrow$ a + b + e + f = 0.005

$P(S_2)=0.008 \rightarrow$ a + b + c + d = 0.008

$P(S_1|D)=0.3 \rightarrow$ (a + e) / (a + c + e + g) = 0.3

$P(S_2|D)=0.25 \rightarrow$ (a + c) / (a + c + e + g) = 0.25

and a + b + c + d + e + f + g + h = 1

So, my question is: are my data sufficient to create a dataset in R, which simulate a real population? The approach I tried was useful in the case of only 1 symptom (How to create a dataset with conditional probability?), as Henry explained. Is it good for the case of two symptoms? Or n-symptoms? Does it exists another approach to simulate this data?

Answer 1

You have (with our trusty old friend Bayes): $$P(D | S1 \& S2) = (P(D \& S1 \& S2)) / (P(S1 \& S2)) = (P(S1 \& S2 | D) P(D)) / (P(S1 \& S2))$$ Given independence and conditional independence of S1 and S2, you know all the parts on the right hand side, so you can calculate the probability you were interested in.

Answer 2

Your question has six equations for eight unknowns, so in theory you need two more such as those additional properties you mention:

• From independence: $\Pr(S_1 \text{ and } S_2) = \Pr(S_1) \Pr(S_2)$ so $a + b =0.00004$

• From conditional independence $\Pr(S_1 \text{ and } S_2 | D) = \Pr(S_1|D) \Pr(S_2|D)$ so $a/(a+c+e+g) = 0.075$

and then use the technique in the previous answer for your sample.

If my calculations are correct then there is a slight problem as you already know $a+c+e+g=0.003$ so these give $a=0.000225$ and thus $b=-0.000185$ so $b$, which should be $\Pr(S_1 \text{ and } S_2 | \text{not } D)$, is not actually a probability.