Is there any bound for the joint probability when the conditional probabilities are difficult to calculate?

When the joint probability for events $A_1,A_2,A_3,A_4$ is desired, the chain rule is used as follows:$$P(A_4, A_3, A_2, A_1) = \mathrm P(A_4 \mid A_3, A_2, A_1)\cdot \mathrm P(A_3 \mid A_2, A_1)\cdot \mathrm P(A_2 \mid A_1)\cdot \mathrm P(A_1)$$ However, I have a problem in which I need the probability of the joint event and calculation of the conditional probabilities is not possible. Is there any approach that I could find an upper or lower bound for this probability? or any sort of approximation, no matter how elementary it is. I only have the probabilities of events alone, i.e. $P(A_4), P(A_3), P(A_2), P(A_1)$.

• It is $P(A_i|A_j)\geq P(A_i)$. Therefore one (rough) lower bound is $P(A_1)\cdot P(A_2)\cdot P(A_3)\cdot P(A_4)$ – callculus Jul 9 '16 at 7:40
• Yes, however this was very far from the actual (I think there is a considerable dependency between them in my problem). I am looking for something more accurate. – asterix Jul 9 '16 at 7:46
• @callculus If $A_i\cap A_j=\varnothing$ then $P(A_i\mid A_j)=0$ and $P(A_i)>0$ is not excluded. – drhab Jul 9 '16 at 7:50
• Do you have any further information ? – callculus Jul 9 '16 at 7:50
• @drhab Yes, I haven´t considered this special case. I thought that all events are dependent. – callculus Jul 9 '16 at 7:54

Without more information upperbound is: $$\min(\Pr(A_1),\Pr(A_2),\Pr(A_3),\Pr(A_4))$$
This is based on $A_1\cap A_2\cap A_3\cap A_4\subseteq A_i$ for $i=1,2,3,4$ together with the fact that $=$ instead of $\subseteq$ is not excluded here.
If e.g. $\Pr(A_1)+\Pr(A_2)\leq1$ then it is not excluded that $A_1\cap A_2=\varnothing$ so in such cases $0$ serves as lower bound. Not quite useful of course. For a useful lower bound more information concerning the events is needed.
• I can only think of things like: $P\left(A_{1}\right)+P\left(A_{2}\right)-1\leq P\left(A_{1}\cap A_{2}\right)\leq\min\left(P\left(A_{1}\right),P\left(A_{2})\right)\right)$ and dividing both sides by $P\left(A_{1}\right)$ gives you something like $\cdots\leq P\left(A_{2}\mid A_{1}\right)\leq\cdots$. – drhab Jul 9 '16 at 8:40