Calculating the distribution functions from two random variables X and Y

Question

I'm currently trying to catch up on Stochastics for university and I'm really stuck on this one although I feel like its not as difficult as I may think:

Let X, Y be random variables. The random variable X takes the values 1, 2 and 3 with probabilities P(X = 1) = 0.5, P(X = 2) = 0.3 and P(X = 3) = 0.2 and Y takes the values 1 and 2 with probabilities P(Y = 1) = 0.7 and P(Y = 2) = 0.3. Moreover, it is known that P(X = 1, Y = 1) = 0.35 and P(X = 3, Y = 1) = 0.2.

a) Calculate the distribution functions of X and Y .

b) Compute the remaining probabilities P(X = i, Y = j) for i ∈ {1, 2, 3} and j ∈ {1, 2}.

Regarding a: I think I just need to see how its done once and than I'll probably get it but my University is not publishing the solutions or the script so my only options are googling and hoping to get it but I really dont :/

Regarding b: I get why P(X = 1, Y = 1) = 0.35 ( because 0.5*0.7 ) but why is P(X = 3, Y = 1) = 0.2 ( 0.2 * 0.7 ? )?

I know its alot and I'm not expecting full on solutions but maybe someone can explain the basic approach to solving the exercise?

I would greatly appreciate it :)

Answer 1

I know its alot and I'm not expecting full on solutions but maybe someone can explain the basic approach to solving the exercise?

a) just calculate the CDF of X and Y thus refer to your textbook.

For example,

$$F_X(x)=0.5\cdot\mathbb{1}_{[1;2)}(x)+0.8\cdot\mathbb{1}_{[2;3)}(x)+\mathbb{1}_{[3;+\infty)}(x)$$

b)

These are your data expressed in a tabular mode

Can you complete the table?

Answer 2

The random variables $X$ and $Y$ must not be independent. If they are not independent then $$P(X=i,Y=j)=P(X=i|Y=j)\cdot P(Y=j)=P(Y=j|X=i)\cdot P(X=i),$$

where $P(X=i|Y=j)\neq P(X=i)$ and $P(Y=j|X=i)\neq P(Y=j)$

In the case of $P(X=3,Y=1)=0.2$ we have

$$P(X=3|Y=1)\cdot P(Y=1)=P(Y=1|X=3)\cdot P(X=3)=0.2$$

$$P(X=3|Y=1)\cdot 0.7=P(Y=1|X=3)\cdot 0.2=0.2$$

Thus $P(X=3|Y=1)=\frac{2}{7}$ and $P(Y=1|X=3)=1$

In the case of bivariate, discrete distribution you can make a table and fill in the given values and then fill in the missing values. The sum of the rows must add to $P(X=i)$ and the sum of the columns must add to $P(Y=j)$

So at $b)$ you have the following table

$$\begin{array}{|m{cm}|m{1cm}|} \hline X/Y& 1 &2 &\textrm{sum} \\ \hline \hline1& 0.35& &0.5\\ \hline 2& &&0.3 \\ \hline 3& 0.2& &0.2 \\\hline \textrm{sum}& 0.7 &0.3 &1\\ \hline \end{array}$$

To fill in the remaining values is simplest algebra.