Prove that the sum and the absolute difference of 2 Bernoulli(0.5) random variables are not independent
Let $X$ and $Y$ be independent $Bernoulli(0.5)$ random variables. Let $W = X + Y$ and $T = |X - Y|$. Show that $W$ and $T$ are not independent.
I know that I have to show that $P(W, T)$ is not equal to $P(W)P(T)$, but finding the joint distribution is hard. Please help.
probability self-study independence bernoulli-distribution
asked Nov 27 '18 at 22:25
MSE
988
add a comment |
Let $X$ and $Y$ be independent $Bernoulli(0.5)$ random variables. Let $W = X + Y$ and $T = |X - Y|$. Show that $W$ and $T$ are not independent.
I know that I have to show that $P(W, T)$ is not equal to $P(W)P(T)$, but finding the joint distribution is hard. Please help.
probability self-study independence bernoulli-distribution
asked Nov 27 '18 at 22:25
MSE
988
Re: "finding the joint distribution is hard:" have you made a table? Label the rows with values of $X$, the columns with values of $Y$, and in the cells put the values of $T,$ $W,$ and the associated probabilities. Collect your results into a new table with rows labeled with $T$ and columns labeled with $W:$ put the total probabilities into the entries. That depicts the entire joint distribution of $(T,W).$ You can then draw your conclusion with a visual inspection. No operation is any more difficult than computing $1/2times 1/2.$
– whuber♦
Nov 27 '18 at 23:19
1
I get it now. I was looking for an elegant, mathematical expression for the joint distribution, but I now realize that I can just enumerate the sample space and the probabilities easily. Thanks, @whuber.
– MSE
Nov 28 '18 at 0:35
add a comment |
Let $X$ and $Y$ be independent $Bernoulli(0.5)$ random variables. Let $W = X + Y$ and $T = |X - Y|$. Show that $W$ and $T$ are not independent.
I know that I have to show that $P(W, T)$ is not equal to $P(W)P(T)$, but finding the joint distribution is hard. Please help.
probability self-study independence bernoulli-distribution
asked Nov 27 '18 at 22:25
MSE
988
Let $X$ and $Y$ be independent $Bernoulli(0.5)$ random variables. Let $W = X + Y$ and $T = |X - Y|$. Show that $W$ and $T$ are not independent.
I know that I have to show that $P(W, T)$ is not equal to $P(W)P(T)$, but finding the joint distribution is hard. Please help.
probability self-study independence bernoulli-distribution
probability self-study independence bernoulli-distribution
asked Nov 27 '18 at 22:25
MSE
988
asked Nov 27 '18 at 22:25
MSE
988
asked Nov 27 '18 at 22:25
MSE
988
asked Nov 27 '18 at 22:25
MSE
988
asked Nov 27 '18 at 22:25
MSE
988
988
Re: "finding the joint distribution is hard:" have you made a table? Label the rows with values of $X$, the columns with values of $Y$, and in the cells put the values of $T,$ $W,$ and the associated probabilities. Collect your results into a new table with rows labeled with $T$ and columns labeled with $W:$ put the total probabilities into the entries. That depicts the entire joint distribution of $(T,W).$ You can then draw your conclusion with a visual inspection. No operation is any more difficult than computing $1/2times 1/2.$
– whuber♦
Nov 27 '18 at 23:19
1
I get it now. I was looking for an elegant, mathematical expression for the joint distribution, but I now realize that I can just enumerate the sample space and the probabilities easily. Thanks, @whuber.
– MSE
Nov 28 '18 at 0:35
add a comment |
Re: "finding the joint distribution is hard:" have you made a table? Label the rows with values of $X$, the columns with values of $Y$, and in the cells put the values of $T,$ $W,$ and the associated probabilities. Collect your results into a new table with rows labeled with $T$ and columns labeled with $W:$ put the total probabilities into the entries. That depicts the entire joint distribution of $(T,W).$ You can then draw your conclusion with a visual inspection. No operation is any more difficult than computing $1/2times 1/2.$
– whuber♦
Nov 27 '18 at 23:19
1
I get it now. I was looking for an elegant, mathematical expression for the joint distribution, but I now realize that I can just enumerate the sample space and the probabilities easily. Thanks, @whuber.
– MSE
Nov 28 '18 at 0:35
Re: "finding the joint distribution is hard:" have you made a table? Label the rows with values of $X$, the columns with values of $Y$, and in the cells put the values of $T,$ $W,$ and the associated probabilities. Collect your results into a new table with rows labeled with $T$ and columns labeled with $W:$ put the total probabilities into the entries. That depicts the entire joint distribution of $(T,W).$ You can then draw your conclusion with a visual inspection. No operation is any more difficult than computing $1/2times 1/2.$
– whuber♦
Nov 27 '18 at 23:19
Re: "finding the joint distribution is hard:" have you made a table? Label the rows with values of $X$, the columns with values of $Y$, and in the cells put the values of $T,$ $W,$ and the associated probabilities. Collect your results into a new table with rows labeled with $T$ and columns labeled with $W:$ put the total probabilities into the entries. That depicts the entire joint distribution of $(T,W).$ You can then draw your conclusion with a visual inspection. No operation is any more difficult than computing $1/2times 1/2.$
– whuber♦
Nov 27 '18 at 23:19
1
1
I get it now. I was looking for an elegant, mathematical expression for the joint distribution, but I now realize that I can just enumerate the sample space and the probabilities easily. Thanks, @whuber.
– MSE
Nov 28 '18 at 0:35
I get it now. I was looking for an elegant, mathematical expression for the joint distribution, but I now realize that I can just enumerate the sample space and the probabilities easily. Thanks, @whuber.
– MSE
Nov 28 '18 at 0:35
add a comment |
2 Answers
2
active
oldest
votes
The product of the marginal distributions is defined on ${0,1,2} times {0,1}$. You can plug in any of the $6$ possible pairs, and get a nonzero number out.
However, the joint density is defined on a smaller space:
$$
{0,0} cup {1,1} cup {2, 0}.
$$
To disprove independence, take any $(w,t)$ pair not in the above, and plug it in to $P(W,T)$ and $P(W)P(T)$. You will see that, for that particular pair:
$$
P(W,T) = 0 neq P(W)P(T).
$$
Alternatively, because you're dealing with a small space, you can just go ahead and compute every probability and just check every possible pair.
answered Nov 27 '18 at 22:48
Taylor
11.6k11745
add a comment |
When T = 0, W = 0 or 2; when T = 1 then W = 1. So T and W are not independent.
See Independence of $X+Y$ and $X-Y$
answered Nov 27 '18 at 22:28
user158565
5,2701318
I want to mark your solution as correct, too! Thanks, @user158565.
– MSE
Nov 28 '18 at 0:39
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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oldest
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active
oldest
votes
The product of the marginal distributions is defined on ${0,1,2} times {0,1}$. You can plug in any of the $6$ possible pairs, and get a nonzero number out.
However, the joint density is defined on a smaller space:
$$
{0,0} cup {1,1} cup {2, 0}.
$$
To disprove independence, take any $(w,t)$ pair not in the above, and plug it in to $P(W,T)$ and $P(W)P(T)$. You will see that, for that particular pair:
$$
P(W,T) = 0 neq P(W)P(T).
$$
Alternatively, because you're dealing with a small space, you can just go ahead and compute every probability and just check every possible pair.
answered Nov 27 '18 at 22:48
Taylor
11.6k11745
add a comment |
The product of the marginal distributions is defined on ${0,1,2} times {0,1}$. You can plug in any of the $6$ possible pairs, and get a nonzero number out.
However, the joint density is defined on a smaller space:
$$
{0,0} cup {1,1} cup {2, 0}.
$$
To disprove independence, take any $(w,t)$ pair not in the above, and plug it in to $P(W,T)$ and $P(W)P(T)$. You will see that, for that particular pair:
$$
P(W,T) = 0 neq P(W)P(T).
$$
Alternatively, because you're dealing with a small space, you can just go ahead and compute every probability and just check every possible pair.
answered Nov 27 '18 at 22:48
Taylor
11.6k11745
add a comment |
The product of the marginal distributions is defined on ${0,1,2} times {0,1}$. You can plug in any of the $6$ possible pairs, and get a nonzero number out.
However, the joint density is defined on a smaller space:
$$
{0,0} cup {1,1} cup {2, 0}.
$$
To disprove independence, take any $(w,t)$ pair not in the above, and plug it in to $P(W,T)$ and $P(W)P(T)$. You will see that, for that particular pair:
$$
P(W,T) = 0 neq P(W)P(T).
$$
Alternatively, because you're dealing with a small space, you can just go ahead and compute every probability and just check every possible pair.
answered Nov 27 '18 at 22:48
Taylor
11.6k11745
The product of the marginal distributions is defined on ${0,1,2} times {0,1}$. You can plug in any of the $6$ possible pairs, and get a nonzero number out.
However, the joint density is defined on a smaller space:
$$
{0,0} cup {1,1} cup {2, 0}.
$$
To disprove independence, take any $(w,t)$ pair not in the above, and plug it in to $P(W,T)$ and $P(W)P(T)$. You will see that, for that particular pair:
$$
P(W,T) = 0 neq P(W)P(T).
$$
Alternatively, because you're dealing with a small space, you can just go ahead and compute every probability and just check every possible pair.
answered Nov 27 '18 at 22:48
Taylor
11.6k11745
answered Nov 27 '18 at 22:48
Taylor
11.6k11745
answered Nov 27 '18 at 22:48
Taylor
11.6k11745
answered Nov 27 '18 at 22:48
Taylor
11.6k11745
11.6k11745
add a comment |
add a comment |
When T = 0, W = 0 or 2; when T = 1 then W = 1. So T and W are not independent.
See Independence of $X+Y$ and $X-Y$
answered Nov 27 '18 at 22:28
user158565
5,2701318
I want to mark your solution as correct, too! Thanks, @user158565.
– MSE
Nov 28 '18 at 0:39
add a comment |
When T = 0, W = 0 or 2; when T = 1 then W = 1. So T and W are not independent.
See Independence of $X+Y$ and $X-Y$
answered Nov 27 '18 at 22:28
user158565
5,2701318
I want to mark your solution as correct, too! Thanks, @user158565.
– MSE
Nov 28 '18 at 0:39
add a comment |
When T = 0, W = 0 or 2; when T = 1 then W = 1. So T and W are not independent.
See Independence of $X+Y$ and $X-Y$
answered Nov 27 '18 at 22:28
user158565
5,2701318
When T = 0, W = 0 or 2; when T = 1 then W = 1. So T and W are not independent.
See Independence of $X+Y$ and $X-Y$
answered Nov 27 '18 at 22:28
user158565
5,2701318
answered Nov 27 '18 at 22:28
user158565
5,2701318
answered Nov 27 '18 at 22:28
user158565
5,2701318
answered Nov 27 '18 at 22:28
user158565
5,2701318
5,2701318
I want to mark your solution as correct, too! Thanks, @user158565.
– MSE
Nov 28 '18 at 0:39
add a comment |
I want to mark your solution as correct, too! Thanks, @user158565.
– MSE
Nov 28 '18 at 0:39
I want to mark your solution as correct, too! Thanks, @user158565.
– MSE
Nov 28 '18 at 0:39
I want to mark your solution as correct, too! Thanks, @user158565.
– MSE
Nov 28 '18 at 0:39
add a comment |
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Re: "finding the joint distribution is hard:" have you made a table? Label the rows with values of $X$, the columns with values of $Y$, and in the cells put the values of $T,$ $W,$ and the associated probabilities. Collect your results into a new table with rows labeled with $T$ and columns labeled with $W:$ put the total probabilities into the entries. That depicts the entire joint distribution of $(T,W).$ You can then draw your conclusion with a visual inspection. No operation is any more difficult than computing $1/2times 1/2.$
– whuber♦
Nov 27 '18 at 23:19
1
I get it now. I was looking for an elegant, mathematical expression for the joint distribution, but I now realize that I can just enumerate the sample space and the probabilities easily. Thanks, @whuber.
– MSE
Nov 28 '18 at 0:35