Probability that X1=0 given that N=1

[DottZakapa]

Homework Statement

Given $X_1,X_2,X_3$ three independent random variables that can assume values 1 ,0 ,-1 with probabilities $\frac 1 4 ,\frac 1 2 ,\frac 1 4$. Let $S=X_1,X_2,X_3$ and let N be the number of$X_i$ assuming value $0$

Relevant Equations

Probability

Given $X_1,X_2,X_3$ three independent random variables that can assume values 1 ,0 ,-1 with probabilities $\frac 1 4 ,\frac 1 2 ,\frac 1 4$. Let $S=X_1,X_2,X_3$ and let N be the number of$X_i$ assuming value $0$.

Knowing that N=1 i have to find what os the probability of $X_1=0$, that is:

$P \left[ X_1=0 | N=1 \right]=$

$P \left[ N=1 \right]= \binom 3 1 * \frac 1 2 *\left(\frac 1 2 \right)^2$

so i can do
$P \left[ X_1=0 | N=1 \right]=\frac {P \left[ X_1=0 \right] P \left[ N=1 | X_1=0 \right]} {P \left[ N=1 \right]}$

now comes the question
in $P \left[ N=1 | X_1=0 \right]$ starts counting ignoring that $X_1 = 0$ so in order to have N=1 there is an other random variable that is equal to 0 and one that is not, $X_1=0, X_2=0, X_3 \neq 0$
or
only $X_1=0$ and the others are not equal zero?

 

[Master1022]

let N be the number of$X_i$ assuming value $0$.

only $X_1=0$ and the others are not equal zero?

Based on the definition of $N$ that you provided, I think your latter scenario is correct (i.e. $X_1 = 0$ and the other two are non-zero)

In terms of the solving, I think Bayes' might overcomplicate it? I think it might be quicker to do the calculation w/o Bayes'. (EDIT: don't change the way you solve it just based on my comment because both methods should yield the same answer - perhaps you use another method as a checking mechanism)

 

[DottZakapa]

Based on the definition of N that you provided, I think your latter scenario is correct (i.e. X1=0 and the other two are non-zero)

In terms of the solving, I think Bayes' might overcomplicate it? I think it might be quicker to do the calculation w/o Bayes'. (EDIT: don't change the way you solve it just based on my comment because both methods should yield the same answer - perhaps you use another method as a checking mechanism)

if$X1=0$ and the other two are non-zero then the result does not coincide with the result (according to the prof correction) which is 1/3.

My solution is :
$P \left[ X_1=0 | N=1 \right]=\frac {P \left[ X_1=0 \right] P \left[ N=1 | X_1=0 \right]} {P \left[ N=1 \right]}=$ $\frac {\frac 1 2 \frac 1 4 \frac 1 4 }{\binom 3 1 \frac 1 2 \left(\frac 1 2 \right)^2}$

while the professor did as follows :
$\frac {\frac 1 2 \frac 1 2 \frac 1 4 }{\binom 3 1 \frac 1 2 \left(\frac 1 2 \right)^2}$

he considered also the second random variable as = to zero and i don't understand why

 

[Master1022]

if$X1=0$ and the other two are non-zero then the result does not coincide with the result (according to the prof correction) which is 1/3.

My solution is :
$P \left[ X_1=0 | N=1 \right]=\frac {P \left[ X_1=0 \right] P \left[ N=1 | X_1=0 \right]} {P \left[ N=1 \right]}=$ $\frac {\frac 1 2 \frac 1 4 \frac 1 4 }{\binom 3 1 \frac 1 2 \left(\frac 1 2 \right)^2}$

can you explain why the numerator is $\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}$ instead of $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}$ as the probability of being non-zero is 1/2?

$\frac {\frac 1 2 \frac 1 2 \frac 1 4 }{\binom 3 1 \frac 1 2 \left(\frac 1 2 \right)^2}$

he considered also the second random variable as = to zero and i don't understand why

I am not sure why he considered a second variable as 0, but will give that more thought afterwards as well. Is it potentially a typo as that expression doesn't = 1/3? I think he meant the numerator to be $( \frac{1}{2} )^3$

I thought the solution would follow:
$$ = \frac{ P(X_1 = 0) P(X_2 \neq 0) P(X_3 \neq 0) }{ P(X_1 = 0) P(X_2 \neq 0) P(X_3 \neq 0) + P(X_1 \neq 0) P(X_2 = 0) P(X_3 \neq 0) + P(X_1 \neq 0) P(X_2 \neq 0) P(X_3 = 0)} $$
which yields $1/3$ as expected

 

[Master1022]

My solution is :
$P \left[ X_1=0 | N=1 \right]=\frac {P \left[ X_1=0 \right] P \left[ N=1 | X_1=0 \right]} {P \left[ N=1 \right]}=$ $\frac {\frac 1 2 \frac 1 4 \frac 1 4 }{\binom 3 1 \frac 1 2 \left(\frac 1 2 \right)^2}$

I think there is an error with you numerator. If you list out all the combinations of $X_1$, $X_2$, and $X_3$ where $X_1 = 0$, there will be 9 combinations. However, not all combinations are equiprobable (as $P(X_i = \pm 1) = 1/4$ each). When I calculated $P(N = 1 | X_1 = 0)$, those are the cases where only $X_1$ is 0, then I get $1/4$ so your numerator would be $\frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8}$ which will yield 1/3

 

[DottZakapa]

i am getting lost
so
tho probability of being 0 is 1/2
probability of being 1 or -1 is 1/4
then, for what concerns
$P \left[ N=1 | X_1=0 \right]$ = $P(X_1=0,X_2=1,X_3=1)+P(X_1=0,X_2=1,X_3=-1)+$
$P(X_1=0,X_2=-1,X_3=1)+P(X_1=0,X_2=-1,X_3=-1)+P(X_1=0,X_2=-1,X_3=-1)$
in which i am taking into account all possible combinations where $X_1=0$.
so it becomes
$\left(\frac 1 2 \frac 1 4 \frac 1 4 \right)*4$
this solves
$P \left[ N=1 | X_1=0 \right]$
but in front of it there also is
$P \left[ X_1=0 \right]$ that is equal to $\frac 1 2$

 

[Master1022]

tho probability of being 0 is 1/2
probability of being 1 or -1 is 1/4

agree

but in front of it there also is
$P \left[ X_1=0 \right]$ that is equal to $\frac 1 2$

agreed

$P \left[ N=1 | X_1=0 \right]$ = $P(X_1=0,X_2=1,X_3=1)+P(X_1=0,X_2=1,X_3=-1)+$
$P(X_1=0,X_2=-1,X_3=1)+P(X_1=0,X_2=-1,X_3=-1)+P(X_1=0,X_2=-1,X_3=-1)$
in which i am taking into account all possible combinations where $X_1=0$.
so it becomes
$\left(\frac 1 2 \frac 1 4 \frac 1 4 \right)*4$
this solves
$P \left[ N=1 | X_1=0 \right]$

I believe this is calculating $P(N = 1 \cap X_1 = 0)$ rather than $P \left[ N=1 | X_1=0 \right]$. You need to divide by $P(X_1 = 0)$ to get
$$ P \left[ N=1 | X_1=0 \right] = \frac{4 \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}}{1/2} = 1/4 $$
Then when you include the extra factor of 1/2 that you mentioned you will get 1/8

Does that make more sense. If not, I think listing out all 9 combinations and finding the probability of each will help calculate this term

 

[Master1022]

When I calculated $P(N = 1 | X_1 = 0)$, those are the cases where only $X_1$ is 0,

Apologies, I can see how this may have been misleading. What I meant here was that I listed out all the combinations where $X_1 = 0$ and found $P(N = 1 | X_1 = 0)$ by doing:
sum of probabilities where ONLY $X_1 = 0$ / sum of probabilities of those 9 combinations

 

[DottZakapa]

$\frac{4 * \frac{1}{2} * \frac{1}{4} * \frac{1}{4}}{1/2} = 1/4$

ok ok, now its all clear. all works.
thank you.

 

[DottZakapa]

Apologies, I can see how this may have been misleading. What I meant here was that I listed out all the combinations where $X_1 = 0$ and found $P(N = 1 | X_1 = 0)$ by doing:
sum of probabilities where ONLY $X_1 = 0$ / sum of probabilities of those 9 combinations

yes i understood that, and being also multiplied by $P(X_1 = 0)$ i remove from the other probability (by dividing because already taken into account ).