# Thread: Easy question about creating a probability distribution

1. ## Easy question about creating a probability distribution

Hi,
I have a question about creating a probability distribution.:

Let's say we have two sets of events with five elements (real numbers) and two random variables X and Y. So, we have one set for each random variable.
In this case, we assign given probabilities to events from these sets.

I am given the following question "find the probability distribution of XY."

Could you explain me what steps I should take in doing this? I tried to multiply each event from X with each event from the other set to find XY and tried to compute the probabilities for each of the 25 events by multiplying the ones from the set of X with the ones from set of Y. However, I was not successful, and I got a total probability over 1. Does this sound too awful? Did I make a mistake here?

I will be really glad, if you could help me to understand what the question means formally and what I should be doing in this situation.

2. Hello,

Consider the values taken by X and the values taken by Y.
Then find the possible values for XY, by multiplying the possibilities.

Then find the probability to get a given value for XY.

for example, if X can have value a with probability p, and Y can have value b with probability q, then P(XY=ab)=P(X=a,Y=b)
And if X and Y are independent, then P(X=a,Y=b)=P(X=a)P(Y=b)

3. Let me try to say it another way: in order to find $P(XY=c)$, you should first find for which values of $c$ this will be non-zero. In other words, find all the products that you can obtain by multiplying the values taken by $X$ and $Y$. If $X$ ranges from 1 to 5 and $Y$ from 0 to 4, then the possible values for the product are 0, 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 20.

Then for all possible value $c$, you have to decompose the event $\{XY=c\}$ into the elementary events that describe all the outcomes for $X$ and $Y$ that would give a product equal to $c$, and sum their probabilities.

For instance, if $X$ and $Y$ have positive integer values, $P(XY=6)=P(X=1, Y=6) +$ $P(X=6,Y=1)+P(X=2,Y=3)+P(X=3,Y=2)$ (of course you can rule out a few cases if you know for instance that $X$ can't be equal to 2).

If you assume that $X,Y$ are independent, then you have $P(X=a, Y=b)=P(X=a)P(Y=b)$ for each term.