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This is my first posting here and I will be grateful for any help.

Say, I have a 4-sided uneven dice. To describe it, we can define a discrete random variable X with the following outcomes:

$P(X=1) = 0.889; \\ P(X=2) = 0.1; \\ P(X=3) = 0.01; \\ P(X=4)=0.001$

Consider a series of three dice rolling. Given that $X$ is a discrete random variable we can compute probability for each set of outcomes (assuming that the order of the outcomes does not matter):

$$P('111') = 0.889\times0.889\times0.889 = 0.7026 \\ \vdots \\ \begin{align} P('123') & = P(x_1=1,x_2=2,x_3=3) + P(x_1=1,x_2=3,x_3=2) + P(x_1=2,x_2=1,x_3=3) + P(x_1=3,x_2=1,x_3=2) + P(x_1=2,x_2=3,x_3=1) + P(x_1=3,x_2=2,x_3=1) \\ & = 6 P(x_1=1,x_2=2,x_3=3) \\ & = 6\times0.889\times0.1\times0.01 = 0.005334 \end{align} \\ \vdots \\ P('444') = 0.001\times0.001\times0.001 = 10^{-9} $$

I hope my logic and calculations above are correct.

My question is, what should I do if I work with a continuous random variable?

For example, $Y \sim N(0,1)$ and I get three outcomes of $Y$: $y_1, y_2, y_3$. Using the Cumulative distribution function, CDF, I can compute the following probability of each outcome:

$$P(Y < y_1) = CDF_Y(y_1) \\ P(Y < y_2) = CDF_Y(y_2) \\ P(Y < y_3) = CDF_Y(y_3)$$

But how do I compute the joint probability of all three outcomes together, $P(Y < y_1, Y < y_2, Y < y_3)$? Am I allowed to simply multiply the three CDFs? Again, the order of the outcomes does not matter.

Thanks for your help,

Ivan

1 Answers 1

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The key criteria is whether or not the random variables are mutually independent.

If they are so, then the product rule for mutually independent events applies.   Whether or not the random variables are continuous or discrete, the joint cumulative distribution function of several mutually independent random variables is the product of their marginal cumulative distribution functions.

$$\mathsf P(Y_1\leq y_1, Y_2\leq y_2,Y_3\leq y_3) ~=~ \mathsf P(Y_1\leq y_1)~\mathsf P(Y_2\leq y_2~)\mathsf P(Y_3\leq y_3)$$

If these random variables are also each continuous–have probability density functions–then the analogous rule applies:

$$f_{Y_1,Y_2,Y_3}(y_1,y_2,y_3)~=~f_{Y_1}(y_1)~f_{Y_2}(y_2)~f_{Y_3}(y_3)$$

But, as stated, that is all only applicable if we have mutual independence.   Elsewise you have to use conditioning.

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    Thanks Graham, yes the events y1, y2 and y3 are mutually independent, so the multiplication should be applicable here. However, I am not sure how to take into account that the order of the events does not matter. I see that the multiplication of the f(y1)*f(y2)*f(y3) to the number of permutations (3! = 6) is a wrong idea (I may get the probability > 1). But I can't figure out another way to take this into account. Or may be in the case of continuous variables (and CDFs) the permutations should be ignored? – 2017-01-26