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Expectation of a function of a random variable using marginal distribution
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This relates to Q5, part (d) here and given below. The solutions are here.
It boils down to the calculation of $E[Y^2]$ in the solution where I wanted to do this by using $f_Y(y)$ and in the answer it is calculated from $f_X,Y(x,y)$.
I used identity: $$ E[g(Y)] = int_-infty^infty g(y) f_Y(y) , dy$$ where here $g(Y) = Y^2$.
And:
$f_Y(y) = y$ for $0 leq y leq 1$ and for $1 < y leq 2$ we have $f_Y(y) = 2 - y $.
So $f_Y(y)$ is a triangle looking like:
I split up the piecewise function and tried calculating (this is where I think I'm perhaps wrong): $$E[Y^2] = int_0^1 g(y) f_Y(y) , dy + int_1^2 g(y) f_Y(y) dy$$
To give:
$$E[Y^2] = int_0^1 y^3 dy + int_1^2 y^2(2-y) , dy$$
but this doesn't give the correct answer - so I'm assuming I've set something up wrong.
For what it's worth, I see why the solution essentially did:
$$ E[Y^2] = int_0^1 int_x^x+1 g(y) , f_X,Y(x,y) , dy , dx $$
with $f_X,Y(x,y) = 1$ as it avoids the piecewise trouble.
My question relates to how to calculate $E[Y^2]$ from $f_Y(y)$. After that I'd happily take advice on when to use $f_Y(y)$ or $f_X,Y(x,y)$ in future.
Any help greatly appreciated, including directing me to further reading.
Mark
probability probability-distributions expectation
asked Aug 2 at 13:37
maw501
336
add a comment |Â
up vote
2
down vote
favorite
This relates to Q5, part (d) here and given below. The solutions are here.
It boils down to the calculation of $E[Y^2]$ in the solution where I wanted to do this by using $f_Y(y)$ and in the answer it is calculated from $f_X,Y(x,y)$.
I used identity: $$ E[g(Y)] = int_-infty^infty g(y) f_Y(y) , dy$$ where here $g(Y) = Y^2$.
And:
$f_Y(y) = y$ for $0 leq y leq 1$ and for $1 < y leq 2$ we have $f_Y(y) = 2 - y $.
So $f_Y(y)$ is a triangle looking like:
I split up the piecewise function and tried calculating (this is where I think I'm perhaps wrong): $$E[Y^2] = int_0^1 g(y) f_Y(y) , dy + int_1^2 g(y) f_Y(y) dy$$
To give:
$$E[Y^2] = int_0^1 y^3 dy + int_1^2 y^2(2-y) , dy$$
but this doesn't give the correct answer - so I'm assuming I've set something up wrong.
For what it's worth, I see why the solution essentially did:
$$ E[Y^2] = int_0^1 int_x^x+1 g(y) , f_X,Y(x,y) , dy , dx $$
with $f_X,Y(x,y) = 1$ as it avoids the piecewise trouble.
My question relates to how to calculate $E[Y^2]$ from $f_Y(y)$. After that I'd happily take advice on when to use $f_Y(y)$ or $f_X,Y(x,y)$ in future.
Any help greatly appreciated, including directing me to further reading.
Mark
probability probability-distributions expectation
asked Aug 2 at 13:37
maw501
336
I haven´t understood yet why you wanted to calculate $E(Y^2)$.
– callculus
Aug 2 at 14:15
To get the variance of $X + Y$ given by: $$mathrmVar , (X + Y) = E[(X + Y)^2] - E[X + Y]^2 = E[X^2] + 2E[XY] + E[Y^2] − (E[X + Y])^2$$
– maw501
Aug 2 at 14:18
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This relates to Q5, part (d) here and given below. The solutions are here.
It boils down to the calculation of $E[Y^2]$ in the solution where I wanted to do this by using $f_Y(y)$ and in the answer it is calculated from $f_X,Y(x,y)$.
I used identity: $$ E[g(Y)] = int_-infty^infty g(y) f_Y(y) , dy$$ where here $g(Y) = Y^2$.
And:
$f_Y(y) = y$ for $0 leq y leq 1$ and for $1 < y leq 2$ we have $f_Y(y) = 2 - y $.
So $f_Y(y)$ is a triangle looking like:
I split up the piecewise function and tried calculating (this is where I think I'm perhaps wrong): $$E[Y^2] = int_0^1 g(y) f_Y(y) , dy + int_1^2 g(y) f_Y(y) dy$$
To give:
$$E[Y^2] = int_0^1 y^3 dy + int_1^2 y^2(2-y) , dy$$
but this doesn't give the correct answer - so I'm assuming I've set something up wrong.
For what it's worth, I see why the solution essentially did:
$$ E[Y^2] = int_0^1 int_x^x+1 g(y) , f_X,Y(x,y) , dy , dx $$
with $f_X,Y(x,y) = 1$ as it avoids the piecewise trouble.
My question relates to how to calculate $E[Y^2]$ from $f_Y(y)$. After that I'd happily take advice on when to use $f_Y(y)$ or $f_X,Y(x,y)$ in future.
Any help greatly appreciated, including directing me to further reading.
Mark
probability probability-distributions expectation
asked Aug 2 at 13:37
maw501
336
This relates to Q5, part (d) here and given below. The solutions are here.
It boils down to the calculation of $E[Y^2]$ in the solution where I wanted to do this by using $f_Y(y)$ and in the answer it is calculated from $f_X,Y(x,y)$.
I used identity: $$ E[g(Y)] = int_-infty^infty g(y) f_Y(y) , dy$$ where here $g(Y) = Y^2$.
And:
$f_Y(y) = y$ for $0 leq y leq 1$ and for $1 < y leq 2$ we have $f_Y(y) = 2 - y $.
So $f_Y(y)$ is a triangle looking like:
I split up the piecewise function and tried calculating (this is where I think I'm perhaps wrong): $$E[Y^2] = int_0^1 g(y) f_Y(y) , dy + int_1^2 g(y) f_Y(y) dy$$
To give:
$$E[Y^2] = int_0^1 y^3 dy + int_1^2 y^2(2-y) , dy$$
but this doesn't give the correct answer - so I'm assuming I've set something up wrong.
For what it's worth, I see why the solution essentially did:
$$ E[Y^2] = int_0^1 int_x^x+1 g(y) , f_X,Y(x,y) , dy , dx $$
with $f_X,Y(x,y) = 1$ as it avoids the piecewise trouble.
My question relates to how to calculate $E[Y^2]$ from $f_Y(y)$. After that I'd happily take advice on when to use $f_Y(y)$ or $f_X,Y(x,y)$ in future.
Any help greatly appreciated, including directing me to further reading.
Mark
probability probability-distributions expectation
asked Aug 2 at 13:37
maw501
336
asked Aug 2 at 13:37
maw501
336
asked Aug 2 at 13:37
maw501
336
asked Aug 2 at 13:37
maw501
336
336
I haven´t understood yet why you wanted to calculate $E(Y^2)$.
– callculus
Aug 2 at 14:15
To get the variance of $X + Y$ given by: $$mathrmVar , (X + Y) = E[(X + Y)^2] - E[X + Y]^2 = E[X^2] + 2E[XY] + E[Y^2] − (E[X + Y])^2$$
– maw501
Aug 2 at 14:18
add a comment |Â
I haven´t understood yet why you wanted to calculate $E(Y^2)$.
– callculus
Aug 2 at 14:15
To get the variance of $X + Y$ given by: $$mathrmVar , (X + Y) = E[(X + Y)^2] - E[X + Y]^2 = E[X^2] + 2E[XY] + E[Y^2] − (E[X + Y])^2$$
– maw501
Aug 2 at 14:18
I haven´t understood yet why you wanted to calculate $E(Y^2)$.
– callculus
Aug 2 at 14:15
I haven´t understood yet why you wanted to calculate $E(Y^2)$.
– callculus
Aug 2 at 14:15
To get the variance of $X + Y$ given by: $$mathrmVar , (X + Y) = E[(X + Y)^2] - E[X + Y]^2 = E[X^2] + 2E[XY] + E[Y^2] − (E[X + Y])^2$$
– maw501
Aug 2 at 14:18
To get the variance of $X + Y$ given by: $$mathrmVar , (X + Y) = E[(X + Y)^2] - E[X + Y]^2 = E[X^2] + 2E[XY] + E[Y^2] − (E[X + Y])^2$$
– maw501
Aug 2 at 14:18
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
I think that you can use both methods. Your method is correct, but I don't understand why you didn't get the right answer, since when I did the calculation you did to find $mathbb E[Y^2]$ from $f_Y(y)$, I got $7/6$, which is the correct answer.
You can use both methods, but which one you pick depends on the information available. In this question, $f_XY(x,y)=1$, so that is easier to work with.
edited Aug 2 at 16:46
callculus
16.2k31427
answered Aug 2 at 14:26
Meeta Jo
1418
Hmmmmmmmm, so it does! Not sure what I was doing before...I would up-vote but don't have the rep. Thanks.
– maw501
Aug 2 at 14:34
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
I think that you can use both methods. Your method is correct, but I don't understand why you didn't get the right answer, since when I did the calculation you did to find $mathbb E[Y^2]$ from $f_Y(y)$, I got $7/6$, which is the correct answer.
You can use both methods, but which one you pick depends on the information available. In this question, $f_XY(x,y)=1$, so that is easier to work with.
edited Aug 2 at 16:46
callculus
16.2k31427
answered Aug 2 at 14:26
Meeta Jo
1418
Hmmmmmmmm, so it does! Not sure what I was doing before...I would up-vote but don't have the rep. Thanks.
– maw501
Aug 2 at 14:34
add a comment |Â
up vote
3
down vote
accepted
I think that you can use both methods. Your method is correct, but I don't understand why you didn't get the right answer, since when I did the calculation you did to find $mathbb E[Y^2]$ from $f_Y(y)$, I got $7/6$, which is the correct answer.
You can use both methods, but which one you pick depends on the information available. In this question, $f_XY(x,y)=1$, so that is easier to work with.
edited Aug 2 at 16:46
callculus
16.2k31427
answered Aug 2 at 14:26
Meeta Jo
1418
Hmmmmmmmm, so it does! Not sure what I was doing before...I would up-vote but don't have the rep. Thanks.
– maw501
Aug 2 at 14:34
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
I think that you can use both methods. Your method is correct, but I don't understand why you didn't get the right answer, since when I did the calculation you did to find $mathbb E[Y^2]$ from $f_Y(y)$, I got $7/6$, which is the correct answer.
You can use both methods, but which one you pick depends on the information available. In this question, $f_XY(x,y)=1$, so that is easier to work with.
edited Aug 2 at 16:46
callculus
16.2k31427
answered Aug 2 at 14:26
Meeta Jo
1418
I think that you can use both methods. Your method is correct, but I don't understand why you didn't get the right answer, since when I did the calculation you did to find $mathbb E[Y^2]$ from $f_Y(y)$, I got $7/6$, which is the correct answer.
You can use both methods, but which one you pick depends on the information available. In this question, $f_XY(x,y)=1$, so that is easier to work with.
edited Aug 2 at 16:46
callculus
16.2k31427
answered Aug 2 at 14:26
Meeta Jo
1418
edited Aug 2 at 16:46
callculus
16.2k31427
edited Aug 2 at 16:46
callculus
16.2k31427
edited Aug 2 at 16:46
callculus
16.2k31427
16.2k31427
answered Aug 2 at 14:26
Meeta Jo
1418
answered Aug 2 at 14:26
Meeta Jo
1418
answered Aug 2 at 14:26
Meeta Jo
1418
1418
Hmmmmmmmm, so it does! Not sure what I was doing before...I would up-vote but don't have the rep. Thanks.
– maw501
Aug 2 at 14:34
add a comment |Â
Hmmmmmmmm, so it does! Not sure what I was doing before...I would up-vote but don't have the rep. Thanks.
– maw501
Aug 2 at 14:34
Hmmmmmmmm, so it does! Not sure what I was doing before...I would up-vote but don't have the rep. Thanks.
– maw501
Aug 2 at 14:34
Hmmmmmmmm, so it does! Not sure what I was doing before...I would up-vote but don't have the rep. Thanks.
– maw501
Aug 2 at 14:34
add a comment |Â
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I haven´t understood yet why you wanted to calculate $E(Y^2)$.
– callculus
Aug 2 at 14:15
To get the variance of $X + Y$ given by: $$mathrmVar , (X + Y) = E[(X + Y)^2] - E[X + Y]^2 = E[X^2] + 2E[XY] + E[Y^2] − (E[X + Y])^2$$
– maw501
Aug 2 at 14:18