Expectation of the function of multiple random variables
$begingroup$
I am working through some basic probability stuff and have a question regarding functions of multiple variables. If I have two random variables $X,Y$ which have some joint probability distribution $P_{XY}(x,y)$ I can obtain expected value of some function $f(x,y)$ by integrating across both variables:
$$
E[f(x,y)] = int int f(x,y) P_{XY}(x,y) dx dy
$$
we can also obtain the expected value of a function of a single variable by following the workings here
$$
E[g(x)] = int g(x) P_{X}(x) dx
$$
If the integration across a function of a single value has meaning I am wondering how we would interpret the integration of a function of multiple variables across a single value
$$
int f(x,y) P_X(x)dx
$$
and wether this expression has any meaning?
probability probability-theory random-variables
asked Jan 8 at 11:48
TomTom
266
$endgroup$
add a comment |
$begingroup$
I am working through some basic probability stuff and have a question regarding functions of multiple variables. If I have two random variables $X,Y$ which have some joint probability distribution $P_{XY}(x,y)$ I can obtain expected value of some function $f(x,y)$ by integrating across both variables:
$$
E[f(x,y)] = int int f(x,y) P_{XY}(x,y) dx dy
$$
we can also obtain the expected value of a function of a single variable by following the workings here
$$
E[g(x)] = int g(x) P_{X}(x) dx
$$
If the integration across a function of a single value has meaning I am wondering how we would interpret the integration of a function of multiple variables across a single value
$$
int f(x,y) P_X(x)dx
$$
and wether this expression has any meaning?
probability probability-theory random-variables
asked Jan 8 at 11:48
TomTom
266
$endgroup$
add a comment |
$begingroup$
I am working through some basic probability stuff and have a question regarding functions of multiple variables. If I have two random variables $X,Y$ which have some joint probability distribution $P_{XY}(x,y)$ I can obtain expected value of some function $f(x,y)$ by integrating across both variables:
$$
E[f(x,y)] = int int f(x,y) P_{XY}(x,y) dx dy
$$
we can also obtain the expected value of a function of a single variable by following the workings here
$$
E[g(x)] = int g(x) P_{X}(x) dx
$$
If the integration across a function of a single value has meaning I am wondering how we would interpret the integration of a function of multiple variables across a single value
$$
int f(x,y) P_X(x)dx
$$
and wether this expression has any meaning?
probability probability-theory random-variables
asked Jan 8 at 11:48
TomTom
266
$endgroup$
I am working through some basic probability stuff and have a question regarding functions of multiple variables. If I have two random variables $X,Y$ which have some joint probability distribution $P_{XY}(x,y)$ I can obtain expected value of some function $f(x,y)$ by integrating across both variables:
$$
E[f(x,y)] = int int f(x,y) P_{XY}(x,y) dx dy
$$
we can also obtain the expected value of a function of a single variable by following the workings here
$$
E[g(x)] = int g(x) P_{X}(x) dx
$$
If the integration across a function of a single value has meaning I am wondering how we would interpret the integration of a function of multiple variables across a single value
$$
int f(x,y) P_X(x)dx
$$
and wether this expression has any meaning?
probability probability-theory random-variables
probability probability-theory random-variables
asked Jan 8 at 11:48
TomTom
266
asked Jan 8 at 11:48
TomTom
266
edited Jan 8 at 11:55
Tom
asked Jan 8 at 11:48
TomTom
266
asked Jan 8 at 11:48
TomTom
266
asked Jan 8 at 11:48
TomTom
266
266
add a comment |
add a comment |
1 Answer
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$begingroup$
The last expression gives you the expectation of $f(X,y)$ where $X$ denotes the random variable and $y$ is fixed.
So defining $Z_y:=f(X,y)$ it gives the expectation of random variable $Z_y$.
answered Jan 8 at 12:01
drhabdrhab
99.7k544130
$endgroup$
$begingroup$
Thank you for your answer! Am i right in thinking that in this case $E[Z_y]$ would be the expectation given a particular value of y and therefore still dependent on the joint distribution?
$endgroup$
– Tom
Jan 8 at 12:10
$begingroup$
There is only dependence wrt the original joint distribution in the sense that the distribution of $X$ is deduced from the joint distribution. That's all. The calculation of $mathbb EZ_y$ does not depend on some condition $Y=y$ as you seem to think.
$endgroup$
– drhab
Jan 8 at 12:18
$begingroup$
$mathbb Ef(X,y)$ and $mathbb E[f(X,Y)mid Y=y]$ are essentially different. If e.g. $X=Y$ then $mathbb E[f(X,Y)mid Y=y]=f(y,y)$, but $mathbb Ef(X,y)$ is calculated on base of the distribution of $X$ and the equality $X=Y$ plays no part in that.
$endgroup$
– drhab
Jan 8 at 12:25
add a comment |
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1 Answer
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$begingroup$
The last expression gives you the expectation of $f(X,y)$ where $X$ denotes the random variable and $y$ is fixed.
So defining $Z_y:=f(X,y)$ it gives the expectation of random variable $Z_y$.
answered Jan 8 at 12:01
drhabdrhab
99.7k544130
$endgroup$
$begingroup$
Thank you for your answer! Am i right in thinking that in this case $E[Z_y]$ would be the expectation given a particular value of y and therefore still dependent on the joint distribution?
$endgroup$
– Tom
Jan 8 at 12:10
$begingroup$
There is only dependence wrt the original joint distribution in the sense that the distribution of $X$ is deduced from the joint distribution. That's all. The calculation of $mathbb EZ_y$ does not depend on some condition $Y=y$ as you seem to think.
$endgroup$
– drhab
Jan 8 at 12:18
$begingroup$
$mathbb Ef(X,y)$ and $mathbb E[f(X,Y)mid Y=y]$ are essentially different. If e.g. $X=Y$ then $mathbb E[f(X,Y)mid Y=y]=f(y,y)$, but $mathbb Ef(X,y)$ is calculated on base of the distribution of $X$ and the equality $X=Y$ plays no part in that.
$endgroup$
– drhab
Jan 8 at 12:25
add a comment |
$begingroup$
The last expression gives you the expectation of $f(X,y)$ where $X$ denotes the random variable and $y$ is fixed.
So defining $Z_y:=f(X,y)$ it gives the expectation of random variable $Z_y$.
answered Jan 8 at 12:01
drhabdrhab
99.7k544130
$endgroup$
$begingroup$
Thank you for your answer! Am i right in thinking that in this case $E[Z_y]$ would be the expectation given a particular value of y and therefore still dependent on the joint distribution?
$endgroup$
– Tom
Jan 8 at 12:10
$begingroup$
There is only dependence wrt the original joint distribution in the sense that the distribution of $X$ is deduced from the joint distribution. That's all. The calculation of $mathbb EZ_y$ does not depend on some condition $Y=y$ as you seem to think.
$endgroup$
– drhab
Jan 8 at 12:18
$begingroup$
$mathbb Ef(X,y)$ and $mathbb E[f(X,Y)mid Y=y]$ are essentially different. If e.g. $X=Y$ then $mathbb E[f(X,Y)mid Y=y]=f(y,y)$, but $mathbb Ef(X,y)$ is calculated on base of the distribution of $X$ and the equality $X=Y$ plays no part in that.
$endgroup$
– drhab
Jan 8 at 12:25
add a comment |
$begingroup$
The last expression gives you the expectation of $f(X,y)$ where $X$ denotes the random variable and $y$ is fixed.
So defining $Z_y:=f(X,y)$ it gives the expectation of random variable $Z_y$.
answered Jan 8 at 12:01
drhabdrhab
99.7k544130
$endgroup$
The last expression gives you the expectation of $f(X,y)$ where $X$ denotes the random variable and $y$ is fixed.
So defining $Z_y:=f(X,y)$ it gives the expectation of random variable $Z_y$.
answered Jan 8 at 12:01
drhabdrhab
99.7k544130
answered Jan 8 at 12:01
drhabdrhab
99.7k544130
answered Jan 8 at 12:01
drhabdrhab
99.7k544130
answered Jan 8 at 12:01
drhabdrhab
99.7k544130
99.7k544130
$begingroup$
Thank you for your answer! Am i right in thinking that in this case $E[Z_y]$ would be the expectation given a particular value of y and therefore still dependent on the joint distribution?
$endgroup$
– Tom
Jan 8 at 12:10
$begingroup$
There is only dependence wrt the original joint distribution in the sense that the distribution of $X$ is deduced from the joint distribution. That's all. The calculation of $mathbb EZ_y$ does not depend on some condition $Y=y$ as you seem to think.
$endgroup$
– drhab
Jan 8 at 12:18
$begingroup$
$mathbb Ef(X,y)$ and $mathbb E[f(X,Y)mid Y=y]$ are essentially different. If e.g. $X=Y$ then $mathbb E[f(X,Y)mid Y=y]=f(y,y)$, but $mathbb Ef(X,y)$ is calculated on base of the distribution of $X$ and the equality $X=Y$ plays no part in that.
$endgroup$
– drhab
Jan 8 at 12:25
add a comment |
$begingroup$
Thank you for your answer! Am i right in thinking that in this case $E[Z_y]$ would be the expectation given a particular value of y and therefore still dependent on the joint distribution?
$endgroup$
– Tom
Jan 8 at 12:10
$begingroup$
There is only dependence wrt the original joint distribution in the sense that the distribution of $X$ is deduced from the joint distribution. That's all. The calculation of $mathbb EZ_y$ does not depend on some condition $Y=y$ as you seem to think.
$endgroup$
– drhab
Jan 8 at 12:18
$begingroup$
$mathbb Ef(X,y)$ and $mathbb E[f(X,Y)mid Y=y]$ are essentially different. If e.g. $X=Y$ then $mathbb E[f(X,Y)mid Y=y]=f(y,y)$, but $mathbb Ef(X,y)$ is calculated on base of the distribution of $X$ and the equality $X=Y$ plays no part in that.
$endgroup$
– drhab
Jan 8 at 12:25
$begingroup$
Thank you for your answer! Am i right in thinking that in this case $E[Z_y]$ would be the expectation given a particular value of y and therefore still dependent on the joint distribution?
$endgroup$
– Tom
Jan 8 at 12:10
$begingroup$
Thank you for your answer! Am i right in thinking that in this case $E[Z_y]$ would be the expectation given a particular value of y and therefore still dependent on the joint distribution?
$endgroup$
– Tom
Jan 8 at 12:10
$begingroup$
There is only dependence wrt the original joint distribution in the sense that the distribution of $X$ is deduced from the joint distribution. That's all. The calculation of $mathbb EZ_y$ does not depend on some condition $Y=y$ as you seem to think.
$endgroup$
– drhab
Jan 8 at 12:18
$begingroup$
There is only dependence wrt the original joint distribution in the sense that the distribution of $X$ is deduced from the joint distribution. That's all. The calculation of $mathbb EZ_y$ does not depend on some condition $Y=y$ as you seem to think.
$endgroup$
– drhab
Jan 8 at 12:18
$begingroup$
$mathbb Ef(X,y)$ and $mathbb E[f(X,Y)mid Y=y]$ are essentially different. If e.g. $X=Y$ then $mathbb E[f(X,Y)mid Y=y]=f(y,y)$, but $mathbb Ef(X,y)$ is calculated on base of the distribution of $X$ and the equality $X=Y$ plays no part in that.
$endgroup$
– drhab
Jan 8 at 12:25
$begingroup$
$mathbb Ef(X,y)$ and $mathbb E[f(X,Y)mid Y=y]$ are essentially different. If e.g. $X=Y$ then $mathbb E[f(X,Y)mid Y=y]=f(y,y)$, but $mathbb Ef(X,y)$ is calculated on base of the distribution of $X$ and the equality $X=Y$ plays no part in that.
$endgroup$
– drhab
Jan 8 at 12:25
add a comment |
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