Mixing
Expectation of a function of two random variables
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
This is a question 2c from here. Solutions here.
Essentially $X$ is an RV of the amount we take to the casino on any given night (less than 40 dollars) and we can show: $$f_X(x) = fracx800$$ and $Y$ is an RV of money we have on leaving the casino, uniform on $[0, 2X]$.
$Z=Y-X$ is an RV of the profit we make on any given night. The question as given is to find $E[Z]$.
Note that: $$f_Z(z) = frac12x$$ and is bounded between $-x$ and $x$.
In the solution we use:
$$f_X,Z(z,x) = f_X(x),f_Z mid X(z mid x)=frac11600$$ and then integrate:
$$ f_Z(z) = int_x=mid z mid^x=40frac11600 , dx $$
to get $E[Z]$.
I have two questions:
- Methodology wise this is fine though we didn't use $Y$ in the solution... I guess this is just a choice to make it simpler given $f_Z(z)$ turns out to be a nice function with terms cancelling?
- I'm confused by the lower bound on the integral of $x=mid z mid$. I get that we can't take less money to the casino than profit we make but I missed this first time around and it doesn't feel natural to define the $x$ limits in terms of $z$ given we have just defined $Z$ in terms of $X$. My guess is that it's due to $f_Z mid X(z mid x)$ only having +ve probability for $x>mid z mid$.
Any clearing up of this confusion (and slight circular feeling I'm having) would be greatly appreciated.
Thanks,
Mark
probability expectation
asked Aug 6 at 8:37
maw501
336
add a comment |Â
up vote
0
down vote
favorite
This is a question 2c from here. Solutions here.
Essentially $X$ is an RV of the amount we take to the casino on any given night (less than 40 dollars) and we can show: $$f_X(x) = fracx800$$ and $Y$ is an RV of money we have on leaving the casino, uniform on $[0, 2X]$.
$Z=Y-X$ is an RV of the profit we make on any given night. The question as given is to find $E[Z]$.
Note that: $$f_Z(z) = frac12x$$ and is bounded between $-x$ and $x$.
In the solution we use:
$$f_X,Z(z,x) = f_X(x),f_Z mid X(z mid x)=frac11600$$ and then integrate:
$$ f_Z(z) = int_x=mid z mid^x=40frac11600 , dx $$
to get $E[Z]$.
I have two questions:
- Methodology wise this is fine though we didn't use $Y$ in the solution... I guess this is just a choice to make it simpler given $f_Z(z)$ turns out to be a nice function with terms cancelling?
- I'm confused by the lower bound on the integral of $x=mid z mid$. I get that we can't take less money to the casino than profit we make but I missed this first time around and it doesn't feel natural to define the $x$ limits in terms of $z$ given we have just defined $Z$ in terms of $X$. My guess is that it's due to $f_Z mid X(z mid x)$ only having +ve probability for $x>mid z mid$.
Any clearing up of this confusion (and slight circular feeling I'm having) would be greatly appreciated.
Thanks,
Mark
probability expectation
asked Aug 6 at 8:37
maw501
336
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This is a question 2c from here. Solutions here.
Essentially $X$ is an RV of the amount we take to the casino on any given night (less than 40 dollars) and we can show: $$f_X(x) = fracx800$$ and $Y$ is an RV of money we have on leaving the casino, uniform on $[0, 2X]$.
$Z=Y-X$ is an RV of the profit we make on any given night. The question as given is to find $E[Z]$.
Note that: $$f_Z(z) = frac12x$$ and is bounded between $-x$ and $x$.
In the solution we use:
$$f_X,Z(z,x) = f_X(x),f_Z mid X(z mid x)=frac11600$$ and then integrate:
$$ f_Z(z) = int_x=mid z mid^x=40frac11600 , dx $$
to get $E[Z]$.
I have two questions:
- Methodology wise this is fine though we didn't use $Y$ in the solution... I guess this is just a choice to make it simpler given $f_Z(z)$ turns out to be a nice function with terms cancelling?
- I'm confused by the lower bound on the integral of $x=mid z mid$. I get that we can't take less money to the casino than profit we make but I missed this first time around and it doesn't feel natural to define the $x$ limits in terms of $z$ given we have just defined $Z$ in terms of $X$. My guess is that it's due to $f_Z mid X(z mid x)$ only having +ve probability for $x>mid z mid$.
Any clearing up of this confusion (and slight circular feeling I'm having) would be greatly appreciated.
Thanks,
Mark
probability expectation
asked Aug 6 at 8:37
maw501
336
This is a question 2c from here. Solutions here.
Essentially $X$ is an RV of the amount we take to the casino on any given night (less than 40 dollars) and we can show: $$f_X(x) = fracx800$$ and $Y$ is an RV of money we have on leaving the casino, uniform on $[0, 2X]$.
$Z=Y-X$ is an RV of the profit we make on any given night. The question as given is to find $E[Z]$.
Note that: $$f_Z(z) = frac12x$$ and is bounded between $-x$ and $x$.
In the solution we use:
$$f_X,Z(z,x) = f_X(x),f_Z mid X(z mid x)=frac11600$$ and then integrate:
$$ f_Z(z) = int_x=mid z mid^x=40frac11600 , dx $$
to get $E[Z]$.
I have two questions:
- Methodology wise this is fine though we didn't use $Y$ in the solution... I guess this is just a choice to make it simpler given $f_Z(z)$ turns out to be a nice function with terms cancelling?
- I'm confused by the lower bound on the integral of $x=mid z mid$. I get that we can't take less money to the casino than profit we make but I missed this first time around and it doesn't feel natural to define the $x$ limits in terms of $z$ given we have just defined $Z$ in terms of $X$. My guess is that it's due to $f_Z mid X(z mid x)$ only having +ve probability for $x>mid z mid$.
Any clearing up of this confusion (and slight circular feeling I'm having) would be greatly appreciated.
Thanks,
Mark
probability expectation
asked Aug 6 at 8:37
maw501
336
edited Aug 6 at 8:43
asked Aug 6 at 8:37
maw501
336
asked Aug 6 at 8:37
maw501
336
asked Aug 6 at 8:37
maw501
336
336
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
1) Methodology wise, you do use $Y$, just a bit indirectly.
2) That is also how you determine the bounds.
Because $Z=Y-X$, and $Ymid Xsim mathcal U[0;2X]$ therefore $Zmid X ~sim~mathcal U[-X;X]$, so ...
$$f_Zmid X(z)=tfrac 12xmathbf 1_zin[-x;x]$$
Because $Z=Y-X$, and $(X,Y)in (x,y): 0leq xleq 40, 0leq yleq 2x$ therefore $(X,Z)in(x,z):0leq xleq 40,-xleq zleq x$, so...
$$(X,Z)in(x,z):-40leq zleq 40~, -zleq x~, zleq xleq 40$$
Putting this together, since $lvert zrvert =max(-z,z)$.... $$beginsplitf_Z(z) &= int_Bbb R f_Zmid X(zmid x)~f_X(x)~mathrm d x \&=int_lvert zrvert^40dfrac 12x~dfrac x800~mathbf 1_zin[-40,40]~mathrm d xendsplit$$
PS: $mathsf E(Z)=mathsf E(mathsf E(Zmid X))=mathsf E(0)$
answered Aug 6 at 9:38
Graham Kemp
80.1k43275
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
1) Methodology wise, you do use $Y$, just a bit indirectly.
2) That is also how you determine the bounds.
Because $Z=Y-X$, and $Ymid Xsim mathcal U[0;2X]$ therefore $Zmid X ~sim~mathcal U[-X;X]$, so ...
$$f_Zmid X(z)=tfrac 12xmathbf 1_zin[-x;x]$$
Because $Z=Y-X$, and $(X,Y)in (x,y): 0leq xleq 40, 0leq yleq 2x$ therefore $(X,Z)in(x,z):0leq xleq 40,-xleq zleq x$, so...
$$(X,Z)in(x,z):-40leq zleq 40~, -zleq x~, zleq xleq 40$$
Putting this together, since $lvert zrvert =max(-z,z)$.... $$beginsplitf_Z(z) &= int_Bbb R f_Zmid X(zmid x)~f_X(x)~mathrm d x \&=int_lvert zrvert^40dfrac 12x~dfrac x800~mathbf 1_zin[-40,40]~mathrm d xendsplit$$
PS: $mathsf E(Z)=mathsf E(mathsf E(Zmid X))=mathsf E(0)$
answered Aug 6 at 9:38
Graham Kemp
80.1k43275
add a comment |Â
up vote
1
down vote
1) Methodology wise, you do use $Y$, just a bit indirectly.
2) That is also how you determine the bounds.
Because $Z=Y-X$, and $Ymid Xsim mathcal U[0;2X]$ therefore $Zmid X ~sim~mathcal U[-X;X]$, so ...
$$f_Zmid X(z)=tfrac 12xmathbf 1_zin[-x;x]$$
Because $Z=Y-X$, and $(X,Y)in (x,y): 0leq xleq 40, 0leq yleq 2x$ therefore $(X,Z)in(x,z):0leq xleq 40,-xleq zleq x$, so...
$$(X,Z)in(x,z):-40leq zleq 40~, -zleq x~, zleq xleq 40$$
Putting this together, since $lvert zrvert =max(-z,z)$.... $$beginsplitf_Z(z) &= int_Bbb R f_Zmid X(zmid x)~f_X(x)~mathrm d x \&=int_lvert zrvert^40dfrac 12x~dfrac x800~mathbf 1_zin[-40,40]~mathrm d xendsplit$$
PS: $mathsf E(Z)=mathsf E(mathsf E(Zmid X))=mathsf E(0)$
answered Aug 6 at 9:38
Graham Kemp
80.1k43275
add a comment |Â
up vote
1
down vote
up vote
1
down vote
1) Methodology wise, you do use $Y$, just a bit indirectly.
2) That is also how you determine the bounds.
Because $Z=Y-X$, and $Ymid Xsim mathcal U[0;2X]$ therefore $Zmid X ~sim~mathcal U[-X;X]$, so ...
$$f_Zmid X(z)=tfrac 12xmathbf 1_zin[-x;x]$$
Because $Z=Y-X$, and $(X,Y)in (x,y): 0leq xleq 40, 0leq yleq 2x$ therefore $(X,Z)in(x,z):0leq xleq 40,-xleq zleq x$, so...
$$(X,Z)in(x,z):-40leq zleq 40~, -zleq x~, zleq xleq 40$$
Putting this together, since $lvert zrvert =max(-z,z)$.... $$beginsplitf_Z(z) &= int_Bbb R f_Zmid X(zmid x)~f_X(x)~mathrm d x \&=int_lvert zrvert^40dfrac 12x~dfrac x800~mathbf 1_zin[-40,40]~mathrm d xendsplit$$
PS: $mathsf E(Z)=mathsf E(mathsf E(Zmid X))=mathsf E(0)$
answered Aug 6 at 9:38
Graham Kemp
80.1k43275
1) Methodology wise, you do use $Y$, just a bit indirectly.
2) That is also how you determine the bounds.
Because $Z=Y-X$, and $Ymid Xsim mathcal U[0;2X]$ therefore $Zmid X ~sim~mathcal U[-X;X]$, so ...
$$f_Zmid X(z)=tfrac 12xmathbf 1_zin[-x;x]$$
Because $Z=Y-X$, and $(X,Y)in (x,y): 0leq xleq 40, 0leq yleq 2x$ therefore $(X,Z)in(x,z):0leq xleq 40,-xleq zleq x$, so...
$$(X,Z)in(x,z):-40leq zleq 40~, -zleq x~, zleq xleq 40$$
Putting this together, since $lvert zrvert =max(-z,z)$.... $$beginsplitf_Z(z) &= int_Bbb R f_Zmid X(zmid x)~f_X(x)~mathrm d x \&=int_lvert zrvert^40dfrac 12x~dfrac x800~mathbf 1_zin[-40,40]~mathrm d xendsplit$$
PS: $mathsf E(Z)=mathsf E(mathsf E(Zmid X))=mathsf E(0)$
answered Aug 6 at 9:38
Graham Kemp
80.1k43275
edited Aug 6 at 9:44
answered Aug 6 at 9:38
Graham Kemp
80.1k43275
answered Aug 6 at 9:38
Graham Kemp
80.1k43275
answered Aug 6 at 9:38
Graham Kemp
80.1k43275
80.1k43275
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2873700%2fexpectation-of-a-function-of-two-random-variables%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password