1. ## Probability Question

Question :

Let $X_1$ and $X_2$ be the joint probability distribution

$. \qquad \qquad \qquad \qquad X_1=0$ $. \qquad \qquad \qquad X_1=1$ $. \qquad \qquad \qquad X_1=2$
$X_2 = 0$ $. \qquad \qquad \qquad 0.1$ $. \qquad \qquad \qquad \qquad 0.4$ $. \qquad \qquad \qquad \qquad 0.1$
$X_2 = 1$ $. \qquad \qquad \qquad 0.2$ $. \qquad \qquad \qquad \qquad 0.2$ $. \qquad \qquad \qquad \qquad 0.0$

i) Find P(X_1 + X_2)
ii) Are X_1 and X_2 independent
iii) P(X_1 = 0 | X_2 = 1)

I need to know how to get the value of $P(X_1 + X_2)$

Here is my work
$
\begin{array}{|c|c|c|} P(X_1 = 0) \ = \ 0.3 \ \ & P(X_1 = 0 |X_2 = 0) \ = \ 0.2 \ \ & P(X_1 = 0 | X_2 = 1) \ = \ 0.5 \\
P(X_1 = 1) \ = \ 0.6 \ \ & P(X_1 = 1 |X_2 = 0) \ = \ 0.7 \ \ & P(X_1 = 1 | X_2 = 1) \ = \ 0.5 \\
P(X_1 = 2) \ = \ 0.1 \ \ & P(X_1 = 2 |X_2 = 0) \ = \ 0.2 \ \ & P(X_1 = 2 | X_2 = 1) \ = \ 0.0 \\
\end{array}
$

And also want to know the if my answers of ii) and iii) are correct or no ?

i) ???

Since $P(X_1|X_2) = P(X_1) \ \therefore \ P(X_1) \ and \ P(X_2)$ are not independent
$
P(X_1 = 0| X_2 = 1) \ = \ 0.5$

2. Im confused. Are they asking you to find the $P(X_1,X_2)$, or literally the value of $X_1$ added to $X_2$, which is a random variable from 0 to 3?

As for independence, they are most surely not. Take the probability of (0,0):

$P(X_1=0,X_2=0) \neq P(X_1=0)P(X_2=0)$

As the probability on the left is 0.1, however the probability on the right is 0.18.

3. Originally Posted by zorro
Question :

Let $X_1$ and $X_2$ be the joint probability distribution

$. \qquad \qquad \qquad \qquad X_1=0$ $. \qquad \qquad \qquad X_1=1$ $. \qquad \qquad \qquad X_1=2$
$X_2 = 0$ $. \qquad \qquad \qquad 0.1$ $. \qquad \qquad \qquad \qquad 0.4$ $. \qquad \qquad \qquad \qquad 0.1$
$X_2 = 1$ $. \qquad \qquad \qquad 0.2$ $. \qquad \qquad \qquad \qquad 0.2$ $. \qquad \qquad \qquad \qquad 0.0$

i) Find P(X_1 + X_2)
ii) Are X_1 and X_2 independent
iii) P(X_1 = 0 | X_2 = 1)

I need to know how to get the value of $P(X_1 + X_2)$

Here is my work
$
\begin{array}{|c|c|c|} P(X_1 = 0) \ = \ 0.3 \ \ & P(X_1 = 0 |X_2 = 0) \ = \ 0.2 \ \ & P(X_1 = 0 | X_2 = 1) \ = \ 0.5 \\
P(X_1 = 1) \ = \ 0.6 \ \ & P(X_1 = 1 |X_2 = 0) \ = \ 0.7 \ \ & P(X_1 = 1 | X_2 = 1) \ = \ 0.5 \\
P(X_1 = 2) \ = \ 0.1 \ \ & P(X_1 = 2 |X_2 = 0) \ = \ 0.2 \ \ & P(X_1 = 2 | X_2 = 1) \ = \ 0.0 \\
\end{array}
$

And also want to know the if my answers of ii) and iii) are correct or no ?

i) ???

Since $P(X_1|X_2) = P(X_1) \ \therefore \ P(X_1) \ and \ P(X_2)$ are not independent
$
P(X_1 = 0| X_2 = 1) \ = \ 0.5$

Your answer for (iii) is correct: 0.2/(0.2 + 0.2 + 0.0) = 1/2.

For (i), I would let $Y = X_1 + X_2$. The possible values for Y are Y = 0, 1, 2, 3. Your job is to calculate Pr(Y = 0), Pr(Y = 1) etc. eg. $\Pr(Y = 1) = \Pr(X_1 = 0 \cap X_2 = 1) + \Pr(X_1 = 1 \cap X_2 = 0) = 0.2 + 0.4 = 0.6$ etc.

For (ii), it's sufficient to show something like $\Pr(X_1 = 0 \cap X_2 = 0) \neq \Pr(X_1 = 0) \cdot \Pr(X_2 = 0)$.

4. ## I wasnt able to understand

Originally Posted by mr fantastic
Your answer for (iii) is correct: 0.2/(0.2 + 0.2 + 0.0) = 1/2.

For (i), I would let $Y = X_1 + X_2$. The possible values for Y are Y = 0, 1, 2, 3. Your job is to calculate Pr(Y = 0), Pr(Y = 1) etc. eg. $\Pr(Y = 1) = \Pr(X_1 = 0 \cap X_2 = 1) + \Pr(X_1 = 1 \cap X_2 = 0) = 0.2 + 0.4 = 0.6$ etc.

For (ii), it's sufficient to show something like $\Pr(X_1 = 0 \cap X_2 = 0) \neq \Pr(X_1 = 0) \cdot \Pr(X_2 = 0)$.

Mr fantastic i wasnt able to understand the answer 1)
could u please tell me why is Y used at all

5. Originally Posted by zorro
Mr fantastic i wasnt able to understand the answer 1)
could u please tell me why is Y used at all
For convenience. I have defined Y to be equal to X1 + X2. If you don't like it, just continue using X1 + X2.

6. $Y$ is the random variable whose values are the values of the random variable $X_1$ and $X_2$ added together. Therefore $Y$ is just the possible combinations of $X_1$ added to $X_2$, and has values that range from 0 to 3.

7. ## Question

Originally Posted by mr fantastic
For convenience. I have defined Y to be equal to X1 + X2. If you don't like it, just continue using X1 + X2.

May i know what is $P(X_1 + X_2)$ is it $P(X_1) + P(X_2)$ or some thing else

In ur prev post i wasnt able to understand this bit

8. Despite your disbelief in the powers of others: As Mr. Fantastic (and myself) said, Y is simply the possible values of X1 (which are 0, 1 or 2), added to the possible values of X2 (which are 0, 1). Therefore Y is a random variable, made from adding two random variables together:

0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 1, 1 + 1= 2, 2 + 0= 2, 2 + 1 = 3;

are all the possible combinations of the two random variables X1 and X2.

It really is that literal.

9. Originally Posted by zorro
May i know what is $P(X_1 + X_2)$ is it $P(X_1) + P(X_2)$ or some thing else

In ur prev post i wasnt able to understand this bit
X1 + X2 is a random variable (which I called Y). $P(X_1 + X_2)$ is the probability of that random variable being equal to the values that it can equal. eg. Pr(X1 + X2 = 0), Pr(X1 + X2 = 1), ..... Pr(X1 + X2 = 3).

10. ## Is this correct?

Originally Posted by mr fantastic
X1 + X2 is a random variable (which I called Y). $P(X_1 + X_2)$ is the probability of that random variable being equal to the values that it can equal. eg. Pr(X1 + X2 = 0), Pr(X1 + X2 = 1), ..... Pr(X1 + X2 = 3).

$P(X_1 + X_2 = 0) = 0.06$
$P(X_1 + X_2 = 1) = 0.6$
$P(X_1 + X_2 = 2) = 0.32$
$P(X_1 + X_2 = 3) = 0.0$
$\therefore$ $P(X_1 + X_2) = 0.98$

Is this right ????

11. The probability that Y takes on some value between 0 and 3 is 1. You are being asked to construct a probability distribution for P(X1+X2).

Also, you should know that 0.98 isn't right simply by the fact that the probability of P(S) (where S is your space) is equal to 1.

12. Originally Posted by zorro
$P(X_1 + X_2 = 0) = 0.06$
$P(X_1 + X_2 = 1) = 0.6$
$P(X_1 + X_2 = 2) = 0.32$
$P(X_1 + X_2 = 3) = 0.0$
$\therefore$ $P(X_1 + X_2) = 0.98$

Is this right ????
$P(X_1 + X_2 = 0) = P(X_1 = 0 \cap X_2 = 0) = 0.1$ straight from the given table.

$P(X_1 + X_2 = 2) = P(X_1 = 1 \cap X_2 = 1) + P(X_1 = 2 \cap X_2 = 0) + = ....$ (get all values from the given table).

etc.