Random Variable: Independent or Dependent?
up vote
0
down vote
favorite
You are given two independent random variables X and Y, where
Pr(X=0) = Pr(X=1) = Pr(Y=0) = Pr(Y=1) = $frac{1}{2}$
Define the random variable Z = X*Y
Are the random variables X and Z independent or dependent?
I know for the random variables to be independent, it should follow that:
P(X=x ^ Z=z) = P(X=x)*P(Z=z)
In this case, taking X and Z at 0:
Pr(X=0 ^ Z=0) = $frac{1}{2}$ * $frac{1}{2}$ = $frac{1}{4}$
I calculated the probability of Z by considering there is a 2/4 probability of Z=0. Not confident about that.
Pr(X)*Pr(Z) = $frac{1}{2}$ * $frac{1}{4}$ = $frac{1}{8}$
For Pr(Z) I got $frac{1}{4}$ by multiplying the probabilities of Y and X at 0.
Since, $frac{1}{4}$ not = $frac{1}{8}$ they are not independent.
Did I approach this correctly?
probability-theory discrete-mathematics probability-distributions random-variables
asked Nov 28 at 0:16
Toby
1207
add a comment |
up vote
0
down vote
favorite
You are given two independent random variables X and Y, where
Pr(X=0) = Pr(X=1) = Pr(Y=0) = Pr(Y=1) = $frac{1}{2}$
Define the random variable Z = X*Y
Are the random variables X and Z independent or dependent?
I know for the random variables to be independent, it should follow that:
P(X=x ^ Z=z) = P(X=x)*P(Z=z)
In this case, taking X and Z at 0:
Pr(X=0 ^ Z=0) = $frac{1}{2}$ * $frac{1}{2}$ = $frac{1}{4}$
I calculated the probability of Z by considering there is a 2/4 probability of Z=0. Not confident about that.
Pr(X)*Pr(Z) = $frac{1}{2}$ * $frac{1}{4}$ = $frac{1}{8}$
For Pr(Z) I got $frac{1}{4}$ by multiplying the probabilities of Y and X at 0.
Since, $frac{1}{4}$ not = $frac{1}{8}$ they are not independent.
Did I approach this correctly?
probability-theory discrete-mathematics probability-distributions random-variables
asked Nov 28 at 0:16
Toby
1207
How did you get $Pr[Z = 0]$ (as well as $Pr[X = 0 land Z = 0]$)? This does not look quite right (although your approach seems fine to me).
– platty
Nov 28 at 0:19
1
$(X,Y) = (0,0), (0,1), (1,0)$ all yield $Z = 0$, so $P(Z = 0) = 3/4$, not $1/2$.
– Bungo
Nov 28 at 0:26
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
You are given two independent random variables X and Y, where
Pr(X=0) = Pr(X=1) = Pr(Y=0) = Pr(Y=1) = $frac{1}{2}$
Define the random variable Z = X*Y
Are the random variables X and Z independent or dependent?
I know for the random variables to be independent, it should follow that:
P(X=x ^ Z=z) = P(X=x)*P(Z=z)
In this case, taking X and Z at 0:
Pr(X=0 ^ Z=0) = $frac{1}{2}$ * $frac{1}{2}$ = $frac{1}{4}$
I calculated the probability of Z by considering there is a 2/4 probability of Z=0. Not confident about that.
Pr(X)*Pr(Z) = $frac{1}{2}$ * $frac{1}{4}$ = $frac{1}{8}$
For Pr(Z) I got $frac{1}{4}$ by multiplying the probabilities of Y and X at 0.
Since, $frac{1}{4}$ not = $frac{1}{8}$ they are not independent.
Did I approach this correctly?
probability-theory discrete-mathematics probability-distributions random-variables
asked Nov 28 at 0:16
Toby
1207
You are given two independent random variables X and Y, where
Pr(X=0) = Pr(X=1) = Pr(Y=0) = Pr(Y=1) = $frac{1}{2}$
Define the random variable Z = X*Y
Are the random variables X and Z independent or dependent?
I know for the random variables to be independent, it should follow that:
P(X=x ^ Z=z) = P(X=x)*P(Z=z)
In this case, taking X and Z at 0:
Pr(X=0 ^ Z=0) = $frac{1}{2}$ * $frac{1}{2}$ = $frac{1}{4}$
I calculated the probability of Z by considering there is a 2/4 probability of Z=0. Not confident about that.
Pr(X)*Pr(Z) = $frac{1}{2}$ * $frac{1}{4}$ = $frac{1}{8}$
For Pr(Z) I got $frac{1}{4}$ by multiplying the probabilities of Y and X at 0.
Since, $frac{1}{4}$ not = $frac{1}{8}$ they are not independent.
Did I approach this correctly?
probability-theory discrete-mathematics probability-distributions random-variables
probability-theory discrete-mathematics probability-distributions random-variables
asked Nov 28 at 0:16
Toby
1207
asked Nov 28 at 0:16
Toby
1207
asked Nov 28 at 0:16
Toby
1207
asked Nov 28 at 0:16
Toby
1207
asked Nov 28 at 0:16
Toby
1207
1207
How did you get $Pr[Z = 0]$ (as well as $Pr[X = 0 land Z = 0]$)? This does not look quite right (although your approach seems fine to me).
– platty
Nov 28 at 0:19
1
$(X,Y) = (0,0), (0,1), (1,0)$ all yield $Z = 0$, so $P(Z = 0) = 3/4$, not $1/2$.
– Bungo
Nov 28 at 0:26
add a comment |
How did you get $Pr[Z = 0]$ (as well as $Pr[X = 0 land Z = 0]$)? This does not look quite right (although your approach seems fine to me).
– platty
Nov 28 at 0:19
1
$(X,Y) = (0,0), (0,1), (1,0)$ all yield $Z = 0$, so $P(Z = 0) = 3/4$, not $1/2$.
– Bungo
Nov 28 at 0:26
How did you get $Pr[Z = 0]$ (as well as $Pr[X = 0 land Z = 0]$)? This does not look quite right (although your approach seems fine to me).
– platty
Nov 28 at 0:19
How did you get $Pr[Z = 0]$ (as well as $Pr[X = 0 land Z = 0]$)? This does not look quite right (although your approach seems fine to me).
– platty
Nov 28 at 0:19
1
1
$(X,Y) = (0,0), (0,1), (1,0)$ all yield $Z = 0$, so $P(Z = 0) = 3/4$, not $1/2$.
– Bungo
Nov 28 at 0:26
$(X,Y) = (0,0), (0,1), (1,0)$ all yield $Z = 0$, so $P(Z = 0) = 3/4$, not $1/2$.
– Bungo
Nov 28 at 0:26
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Check the distribution of $Z$. First, it can only take the values $0$ and $1$. But as ${Z=1}$ is equivalent to ${X=1wedge Y=1}$, then $P(Z=1)=frac12cdotfrac12=frac14$ and so $P(Z=0)=frac34$.
Also check that
$$P(Z=0|X=0)=1,$$
and since $P(Z=0)neq 1$, this implies $X$ and $Z$ are not independent, because knowing that $X=0$ modifies the probability of $Z=0$, that is
$$P(Z=0|X=0)neq P(Z=0).$$
answered Nov 28 at 0:27
Alejandro Nasif Salum
3,989117
I'm a little confused on how to calculate P(X=1 ^ Z=1). I know P(Z) = 1/4 based off that there is only one instance of (x,y)=(1,1) from 4.
– Toby
Nov 28 at 0:37
Well, actually you should use conditional probability. $$P(X=1,Z=1)=P(Z=1|X=1)cdot P(X=1).$$
– Alejandro Nasif Salum
Nov 28 at 0:40
Do I calculate the conditional probability like this: We need to find Z=1 with event X=1. So, (x,y): (0.0), (0,1), (1,0), (1,1). So there is a 1/4 probability that Z=1 with X=1? So, P(X=1 ^ Z=1) = 1/4 * 1/2 = 1/8? This doesn't seem right as it then becomes independent
– Toby
Nov 28 at 0:53
1
No, if $X=1$, the only possible cases are two: $(1,0)$ and $(1,1)$. And only the last one gives $Z=1$. So $P(Z=1|X=1)=frac12$.
– Alejandro Nasif Salum
Nov 28 at 0:57
add a comment |
up vote
0
down vote
Hint: $XY$ cannot be $1$ when $X$ is $0$. However $X$ may be $0$ and $XY$ may be $1$.
answered Nov 28 at 0:27
Graham Kemp
84.7k43378
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Check the distribution of $Z$. First, it can only take the values $0$ and $1$. But as ${Z=1}$ is equivalent to ${X=1wedge Y=1}$, then $P(Z=1)=frac12cdotfrac12=frac14$ and so $P(Z=0)=frac34$.
Also check that
$$P(Z=0|X=0)=1,$$
and since $P(Z=0)neq 1$, this implies $X$ and $Z$ are not independent, because knowing that $X=0$ modifies the probability of $Z=0$, that is
$$P(Z=0|X=0)neq P(Z=0).$$
answered Nov 28 at 0:27
Alejandro Nasif Salum
3,989117
I'm a little confused on how to calculate P(X=1 ^ Z=1). I know P(Z) = 1/4 based off that there is only one instance of (x,y)=(1,1) from 4.
– Toby
Nov 28 at 0:37
Well, actually you should use conditional probability. $$P(X=1,Z=1)=P(Z=1|X=1)cdot P(X=1).$$
– Alejandro Nasif Salum
Nov 28 at 0:40
Do I calculate the conditional probability like this: We need to find Z=1 with event X=1. So, (x,y): (0.0), (0,1), (1,0), (1,1). So there is a 1/4 probability that Z=1 with X=1? So, P(X=1 ^ Z=1) = 1/4 * 1/2 = 1/8? This doesn't seem right as it then becomes independent
– Toby
Nov 28 at 0:53
1
No, if $X=1$, the only possible cases are two: $(1,0)$ and $(1,1)$. And only the last one gives $Z=1$. So $P(Z=1|X=1)=frac12$.
– Alejandro Nasif Salum
Nov 28 at 0:57
add a comment |
up vote
1
down vote
accepted
Check the distribution of $Z$. First, it can only take the values $0$ and $1$. But as ${Z=1}$ is equivalent to ${X=1wedge Y=1}$, then $P(Z=1)=frac12cdotfrac12=frac14$ and so $P(Z=0)=frac34$.
Also check that
$$P(Z=0|X=0)=1,$$
and since $P(Z=0)neq 1$, this implies $X$ and $Z$ are not independent, because knowing that $X=0$ modifies the probability of $Z=0$, that is
$$P(Z=0|X=0)neq P(Z=0).$$
answered Nov 28 at 0:27
Alejandro Nasif Salum
3,989117
I'm a little confused on how to calculate P(X=1 ^ Z=1). I know P(Z) = 1/4 based off that there is only one instance of (x,y)=(1,1) from 4.
– Toby
Nov 28 at 0:37
Well, actually you should use conditional probability. $$P(X=1,Z=1)=P(Z=1|X=1)cdot P(X=1).$$
– Alejandro Nasif Salum
Nov 28 at 0:40
Do I calculate the conditional probability like this: We need to find Z=1 with event X=1. So, (x,y): (0.0), (0,1), (1,0), (1,1). So there is a 1/4 probability that Z=1 with X=1? So, P(X=1 ^ Z=1) = 1/4 * 1/2 = 1/8? This doesn't seem right as it then becomes independent
– Toby
Nov 28 at 0:53
1
No, if $X=1$, the only possible cases are two: $(1,0)$ and $(1,1)$. And only the last one gives $Z=1$. So $P(Z=1|X=1)=frac12$.
– Alejandro Nasif Salum
Nov 28 at 0:57
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Check the distribution of $Z$. First, it can only take the values $0$ and $1$. But as ${Z=1}$ is equivalent to ${X=1wedge Y=1}$, then $P(Z=1)=frac12cdotfrac12=frac14$ and so $P(Z=0)=frac34$.
Also check that
$$P(Z=0|X=0)=1,$$
and since $P(Z=0)neq 1$, this implies $X$ and $Z$ are not independent, because knowing that $X=0$ modifies the probability of $Z=0$, that is
$$P(Z=0|X=0)neq P(Z=0).$$
answered Nov 28 at 0:27
Alejandro Nasif Salum
3,989117
Check the distribution of $Z$. First, it can only take the values $0$ and $1$. But as ${Z=1}$ is equivalent to ${X=1wedge Y=1}$, then $P(Z=1)=frac12cdotfrac12=frac14$ and so $P(Z=0)=frac34$.
Also check that
$$P(Z=0|X=0)=1,$$
and since $P(Z=0)neq 1$, this implies $X$ and $Z$ are not independent, because knowing that $X=0$ modifies the probability of $Z=0$, that is
$$P(Z=0|X=0)neq P(Z=0).$$
answered Nov 28 at 0:27
Alejandro Nasif Salum
3,989117
edited Nov 28 at 0:33
answered Nov 28 at 0:27
Alejandro Nasif Salum
3,989117
answered Nov 28 at 0:27
Alejandro Nasif Salum
3,989117
answered Nov 28 at 0:27
Alejandro Nasif Salum
3,989117
3,989117
I'm a little confused on how to calculate P(X=1 ^ Z=1). I know P(Z) = 1/4 based off that there is only one instance of (x,y)=(1,1) from 4.
– Toby
Nov 28 at 0:37
Well, actually you should use conditional probability. $$P(X=1,Z=1)=P(Z=1|X=1)cdot P(X=1).$$
– Alejandro Nasif Salum
Nov 28 at 0:40
Do I calculate the conditional probability like this: We need to find Z=1 with event X=1. So, (x,y): (0.0), (0,1), (1,0), (1,1). So there is a 1/4 probability that Z=1 with X=1? So, P(X=1 ^ Z=1) = 1/4 * 1/2 = 1/8? This doesn't seem right as it then becomes independent
– Toby
Nov 28 at 0:53
1
No, if $X=1$, the only possible cases are two: $(1,0)$ and $(1,1)$. And only the last one gives $Z=1$. So $P(Z=1|X=1)=frac12$.
– Alejandro Nasif Salum
Nov 28 at 0:57
add a comment |
I'm a little confused on how to calculate P(X=1 ^ Z=1). I know P(Z) = 1/4 based off that there is only one instance of (x,y)=(1,1) from 4.
– Toby
Nov 28 at 0:37
Well, actually you should use conditional probability. $$P(X=1,Z=1)=P(Z=1|X=1)cdot P(X=1).$$
– Alejandro Nasif Salum
Nov 28 at 0:40
Do I calculate the conditional probability like this: We need to find Z=1 with event X=1. So, (x,y): (0.0), (0,1), (1,0), (1,1). So there is a 1/4 probability that Z=1 with X=1? So, P(X=1 ^ Z=1) = 1/4 * 1/2 = 1/8? This doesn't seem right as it then becomes independent
– Toby
Nov 28 at 0:53
1
No, if $X=1$, the only possible cases are two: $(1,0)$ and $(1,1)$. And only the last one gives $Z=1$. So $P(Z=1|X=1)=frac12$.
– Alejandro Nasif Salum
Nov 28 at 0:57
I'm a little confused on how to calculate P(X=1 ^ Z=1). I know P(Z) = 1/4 based off that there is only one instance of (x,y)=(1,1) from 4.
– Toby
Nov 28 at 0:37
I'm a little confused on how to calculate P(X=1 ^ Z=1). I know P(Z) = 1/4 based off that there is only one instance of (x,y)=(1,1) from 4.
– Toby
Nov 28 at 0:37
Well, actually you should use conditional probability. $$P(X=1,Z=1)=P(Z=1|X=1)cdot P(X=1).$$
– Alejandro Nasif Salum
Nov 28 at 0:40
Well, actually you should use conditional probability. $$P(X=1,Z=1)=P(Z=1|X=1)cdot P(X=1).$$
– Alejandro Nasif Salum
Nov 28 at 0:40
Do I calculate the conditional probability like this: We need to find Z=1 with event X=1. So, (x,y): (0.0), (0,1), (1,0), (1,1). So there is a 1/4 probability that Z=1 with X=1? So, P(X=1 ^ Z=1) = 1/4 * 1/2 = 1/8? This doesn't seem right as it then becomes independent
– Toby
Nov 28 at 0:53
Do I calculate the conditional probability like this: We need to find Z=1 with event X=1. So, (x,y): (0.0), (0,1), (1,0), (1,1). So there is a 1/4 probability that Z=1 with X=1? So, P(X=1 ^ Z=1) = 1/4 * 1/2 = 1/8? This doesn't seem right as it then becomes independent
– Toby
Nov 28 at 0:53
1
1
No, if $X=1$, the only possible cases are two: $(1,0)$ and $(1,1)$. And only the last one gives $Z=1$. So $P(Z=1|X=1)=frac12$.
– Alejandro Nasif Salum
Nov 28 at 0:57
No, if $X=1$, the only possible cases are two: $(1,0)$ and $(1,1)$. And only the last one gives $Z=1$. So $P(Z=1|X=1)=frac12$.
– Alejandro Nasif Salum
Nov 28 at 0:57
add a comment |
up vote
0
down vote
Hint: $XY$ cannot be $1$ when $X$ is $0$. However $X$ may be $0$ and $XY$ may be $1$.
answered Nov 28 at 0:27
Graham Kemp
84.7k43378
add a comment |
up vote
0
down vote
Hint: $XY$ cannot be $1$ when $X$ is $0$. However $X$ may be $0$ and $XY$ may be $1$.
answered Nov 28 at 0:27
Graham Kemp
84.7k43378
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint: $XY$ cannot be $1$ when $X$ is $0$. However $X$ may be $0$ and $XY$ may be $1$.
answered Nov 28 at 0:27
Graham Kemp
84.7k43378
Hint: $XY$ cannot be $1$ when $X$ is $0$. However $X$ may be $0$ and $XY$ may be $1$.
answered Nov 28 at 0:27
Graham Kemp
84.7k43378
answered Nov 28 at 0:27
Graham Kemp
84.7k43378
answered Nov 28 at 0:27
Graham Kemp
84.7k43378
answered Nov 28 at 0:27
Graham Kemp
84.7k43378
84.7k43378
add a comment |
add a comment |
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How did you get $Pr[Z = 0]$ (as well as $Pr[X = 0 land Z = 0]$)? This does not look quite right (although your approach seems fine to me).
– platty
Nov 28 at 0:19
1
$(X,Y) = (0,0), (0,1), (1,0)$ all yield $Z = 0$, so $P(Z = 0) = 3/4$, not $1/2$.
– Bungo
Nov 28 at 0:26