Joint probability distribution from a linear combination [closed]
I have two independent random variables $X$ and $Y$ which summed together give another random variable $X + Y = Z$. I also have the probability distributions for both (i.e $f_X(x)$ and $f_Y(y)$ ).
Given that I know $X$ and $Y$ are independent, their probability distributions is it possible to work out the joint distribution of $X$ and $Z$ ( $f_{XZ} (x,z)$) ?
probability-theory statistics
asked Nov 24 at 11:52
Tom
225
closed as off-topic by Kavi Rama Murthy, John B, user10354138, Cyclohexanol., Cesareo Dec 1 at 10:14
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I have two independent random variables $X$ and $Y$ which summed together give another random variable $X + Y = Z$. I also have the probability distributions for both (i.e $f_X(x)$ and $f_Y(y)$ ).
Given that I know $X$ and $Y$ are independent, their probability distributions is it possible to work out the joint distribution of $X$ and $Z$ ( $f_{XZ} (x,z)$) ?
probability-theory statistics
asked Nov 24 at 11:52
Tom
225
closed as off-topic by Kavi Rama Murthy, John B, user10354138, Cyclohexanol., Cesareo Dec 1 at 10:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Kavi Rama Murthy, John B, user10354138, Cyclohexanol., Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
I have two independent random variables $X$ and $Y$ which summed together give another random variable $X + Y = Z$. I also have the probability distributions for both (i.e $f_X(x)$ and $f_Y(y)$ ).
Given that I know $X$ and $Y$ are independent, their probability distributions is it possible to work out the joint distribution of $X$ and $Z$ ( $f_{XZ} (x,z)$) ?
probability-theory statistics
asked Nov 24 at 11:52
Tom
225
I have two independent random variables $X$ and $Y$ which summed together give another random variable $X + Y = Z$. I also have the probability distributions for both (i.e $f_X(x)$ and $f_Y(y)$ ).
Given that I know $X$ and $Y$ are independent, their probability distributions is it possible to work out the joint distribution of $X$ and $Z$ ( $f_{XZ} (x,z)$) ?
probability-theory statistics
probability-theory statistics
asked Nov 24 at 11:52
Tom
225
asked Nov 24 at 11:52
Tom
225
asked Nov 24 at 11:52
Tom
225
asked Nov 24 at 11:52
Tom
225
asked Nov 24 at 11:52
Tom
225
225
closed as off-topic by Kavi Rama Murthy, John B, user10354138, Cyclohexanol., Cesareo Dec 1 at 10:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Kavi Rama Murthy, John B, user10354138, Cyclohexanol., Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Kavi Rama Murthy, John B, user10354138, Cyclohexanol., Cesareo Dec 1 at 10:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Kavi Rama Murthy, John B, user10354138, Cyclohexanol., Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
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Since $X, Y$ are independent, we know that $f_{X,Y}(x, y) = f_X(x) f_Y(y)$. In this case the change of variables is straightforward, just replace $y$ with $z - x$. So the joint density would be $f_{X,Z}(x, z) = f_X(x) f_Y(z - x)$. You can find out more about changing variables in a probability density function on this Wikipedia page.
answered Nov 24 at 21:00
Alex
534312
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Since $X, Y$ are independent, we know that $f_{X,Y}(x, y) = f_X(x) f_Y(y)$. In this case the change of variables is straightforward, just replace $y$ with $z - x$. So the joint density would be $f_{X,Z}(x, z) = f_X(x) f_Y(z - x)$. You can find out more about changing variables in a probability density function on this Wikipedia page.
answered Nov 24 at 21:00
Alex
534312
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Since $X, Y$ are independent, we know that $f_{X,Y}(x, y) = f_X(x) f_Y(y)$. In this case the change of variables is straightforward, just replace $y$ with $z - x$. So the joint density would be $f_{X,Z}(x, z) = f_X(x) f_Y(z - x)$. You can find out more about changing variables in a probability density function on this Wikipedia page.
answered Nov 24 at 21:00
Alex
534312
add a comment |
Since $X, Y$ are independent, we know that $f_{X,Y}(x, y) = f_X(x) f_Y(y)$. In this case the change of variables is straightforward, just replace $y$ with $z - x$. So the joint density would be $f_{X,Z}(x, z) = f_X(x) f_Y(z - x)$. You can find out more about changing variables in a probability density function on this Wikipedia page.
answered Nov 24 at 21:00
Alex
534312
Since $X, Y$ are independent, we know that $f_{X,Y}(x, y) = f_X(x) f_Y(y)$. In this case the change of variables is straightforward, just replace $y$ with $z - x$. So the joint density would be $f_{X,Z}(x, z) = f_X(x) f_Y(z - x)$. You can find out more about changing variables in a probability density function on this Wikipedia page.
answered Nov 24 at 21:00
Alex
534312
answered Nov 24 at 21:00
Alex
534312
answered Nov 24 at 21:00
Alex
534312
answered Nov 24 at 21:00
Alex
534312
534312
add a comment |
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