Mixing
calculating $E[3xy]$ in die throw - mostly about the way you perform the calculation
$begingroup$
I don't understand the calculation given in the answers for finding mean of a die throw.
edited because of errata in book: experiment: you throw a fair die. we define X: to be the result of the throw, and also defining Y=0 if the result is even and Y=1 if it's odd
Find: $E[3xy]$.
My attempt:
$E[3xy] = 3E[xy]$. so to find $E[xy]$ I have to calculate $$e[xy]=sum_xsum_y xyP(x=i)P(y=j)$$
(is it correct?), or can I calculate it otherwise? so the calculation is, when $(n(S) = 1/6), n(S) cdot 3 cdot (1+3+5) cdots $
and my question is: do we calculate it by the result of a die?
I mean we don't care if its $1$ or $3$, we care that the probability is the same, i.e $1/6$.
what is the difference between the mean of e[x] and e[y]? that we calculate in e[y] only the odd values and e[x] = 3.5?
on a side question, if possible: can we calculate $e[xy]$ using the marginal probability function(using only one variable, $x$ or $y$ for instance)?
would really appreciate an explanation if possible.
thank you
probability probability-distributions
asked Jan 29 at 9:17
q123q123
75
$endgroup$
add a comment |
$begingroup$
I don't understand the calculation given in the answers for finding mean of a die throw.
edited because of errata in book: experiment: you throw a fair die. we define X: to be the result of the throw, and also defining Y=0 if the result is even and Y=1 if it's odd
Find: $E[3xy]$.
My attempt:
$E[3xy] = 3E[xy]$. so to find $E[xy]$ I have to calculate $$e[xy]=sum_xsum_y xyP(x=i)P(y=j)$$
(is it correct?), or can I calculate it otherwise? so the calculation is, when $(n(S) = 1/6), n(S) cdot 3 cdot (1+3+5) cdots $
and my question is: do we calculate it by the result of a die?
I mean we don't care if its $1$ or $3$, we care that the probability is the same, i.e $1/6$.
what is the difference between the mean of e[x] and e[y]? that we calculate in e[y] only the odd values and e[x] = 3.5?
on a side question, if possible: can we calculate $e[xy]$ using the marginal probability function(using only one variable, $x$ or $y$ for instance)?
would really appreciate an explanation if possible.
thank you
probability probability-distributions
asked Jan 29 at 9:17
q123q123
75
$endgroup$
$begingroup$
@SatishRamanathan edited. does it make more sense now?
$endgroup$
– q123
Jan 29 at 9:28
$begingroup$
I don't really get what the values for $X$ and $Y$ are. What value does $X$ have for an odd result? Same question for $Y$ on an even result.
$endgroup$
– Harnak
Jan 29 at 9:40
$begingroup$
bah, i was wonering that the same until i looked at the errata. now it makes more sense. editing again
$endgroup$
– q123
Jan 29 at 9:43
$begingroup$
i got that $e[x] = 16*(1+2+...+6) = 3.5$ and $e[y]= 1*(1+3+5)*1/6 + 0*16*(2+4+6) = 9/6=1.5$. i think that e[y] should be 0.5, because of the table, but i don't know how to show it mathmatically
$endgroup$
– q123
Jan 29 at 10:00
add a comment |
$begingroup$
I don't understand the calculation given in the answers for finding mean of a die throw.
edited because of errata in book: experiment: you throw a fair die. we define X: to be the result of the throw, and also defining Y=0 if the result is even and Y=1 if it's odd
Find: $E[3xy]$.
My attempt:
$E[3xy] = 3E[xy]$. so to find $E[xy]$ I have to calculate $$e[xy]=sum_xsum_y xyP(x=i)P(y=j)$$
(is it correct?), or can I calculate it otherwise? so the calculation is, when $(n(S) = 1/6), n(S) cdot 3 cdot (1+3+5) cdots $
and my question is: do we calculate it by the result of a die?
I mean we don't care if its $1$ or $3$, we care that the probability is the same, i.e $1/6$.
what is the difference between the mean of e[x] and e[y]? that we calculate in e[y] only the odd values and e[x] = 3.5?
on a side question, if possible: can we calculate $e[xy]$ using the marginal probability function(using only one variable, $x$ or $y$ for instance)?
would really appreciate an explanation if possible.
thank you
probability probability-distributions
asked Jan 29 at 9:17
q123q123
75
$endgroup$
I don't understand the calculation given in the answers for finding mean of a die throw.
edited because of errata in book: experiment: you throw a fair die. we define X: to be the result of the throw, and also defining Y=0 if the result is even and Y=1 if it's odd
Find: $E[3xy]$.
My attempt:
$E[3xy] = 3E[xy]$. so to find $E[xy]$ I have to calculate $$e[xy]=sum_xsum_y xyP(x=i)P(y=j)$$
(is it correct?), or can I calculate it otherwise? so the calculation is, when $(n(S) = 1/6), n(S) cdot 3 cdot (1+3+5) cdots $
and my question is: do we calculate it by the result of a die?
I mean we don't care if its $1$ or $3$, we care that the probability is the same, i.e $1/6$.
what is the difference between the mean of e[x] and e[y]? that we calculate in e[y] only the odd values and e[x] = 3.5?
on a side question, if possible: can we calculate $e[xy]$ using the marginal probability function(using only one variable, $x$ or $y$ for instance)?
would really appreciate an explanation if possible.
thank you
probability probability-distributions
probability probability-distributions
asked Jan 29 at 9:17
q123q123
75
asked Jan 29 at 9:17
q123q123
75
edited Jan 29 at 9:49
q123
asked Jan 29 at 9:17
q123q123
75
asked Jan 29 at 9:17
q123q123
75
asked Jan 29 at 9:17
q123q123
75
75
$begingroup$
@SatishRamanathan edited. does it make more sense now?
$endgroup$
– q123
Jan 29 at 9:28
$begingroup$
I don't really get what the values for $X$ and $Y$ are. What value does $X$ have for an odd result? Same question for $Y$ on an even result.
$endgroup$
– Harnak
Jan 29 at 9:40
$begingroup$
bah, i was wonering that the same until i looked at the errata. now it makes more sense. editing again
$endgroup$
– q123
Jan 29 at 9:43
$begingroup$
i got that $e[x] = 16*(1+2+...+6) = 3.5$ and $e[y]= 1*(1+3+5)*1/6 + 0*16*(2+4+6) = 9/6=1.5$. i think that e[y] should be 0.5, because of the table, but i don't know how to show it mathmatically
$endgroup$
– q123
Jan 29 at 10:00
add a comment |
$begingroup$
@SatishRamanathan edited. does it make more sense now?
$endgroup$
– q123
Jan 29 at 9:28
$begingroup$
I don't really get what the values for $X$ and $Y$ are. What value does $X$ have for an odd result? Same question for $Y$ on an even result.
$endgroup$
– Harnak
Jan 29 at 9:40
$begingroup$
bah, i was wonering that the same until i looked at the errata. now it makes more sense. editing again
$endgroup$
– q123
Jan 29 at 9:43
$begingroup$
i got that $e[x] = 16*(1+2+...+6) = 3.5$ and $e[y]= 1*(1+3+5)*1/6 + 0*16*(2+4+6) = 9/6=1.5$. i think that e[y] should be 0.5, because of the table, but i don't know how to show it mathmatically
$endgroup$
– q123
Jan 29 at 10:00
$begingroup$
@SatishRamanathan edited. does it make more sense now?
$endgroup$
– q123
Jan 29 at 9:28
$begingroup$
@SatishRamanathan edited. does it make more sense now?
$endgroup$
– q123
Jan 29 at 9:28
$begingroup$
I don't really get what the values for $X$ and $Y$ are. What value does $X$ have for an odd result? Same question for $Y$ on an even result.
$endgroup$
– Harnak
Jan 29 at 9:40
$begingroup$
I don't really get what the values for $X$ and $Y$ are. What value does $X$ have for an odd result? Same question for $Y$ on an even result.
$endgroup$
– Harnak
Jan 29 at 9:40
$begingroup$
bah, i was wonering that the same until i looked at the errata. now it makes more sense. editing again
$endgroup$
– q123
Jan 29 at 9:43
$begingroup$
bah, i was wonering that the same until i looked at the errata. now it makes more sense. editing again
$endgroup$
– q123
Jan 29 at 9:43
$begingroup$
i got that $e[x] = 16*(1+2+...+6) = 3.5$ and $e[y]= 1*(1+3+5)*1/6 + 0*16*(2+4+6) = 9/6=1.5$. i think that e[y] should be 0.5, because of the table, but i don't know how to show it mathmatically
$endgroup$
– q123
Jan 29 at 10:00
$begingroup$
i got that $e[x] = 16*(1+2+...+6) = 3.5$ and $e[y]= 1*(1+3+5)*1/6 + 0*16*(2+4+6) = 9/6=1.5$. i think that e[y] should be 0.5, because of the table, but i don't know how to show it mathmatically
$endgroup$
– q123
Jan 29 at 10:00
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is not correct to state that $mathbb E[XY]=sum_xsum_yxyP(X=x)P(Y=y)$.
It is correct to state that: $$mathbb E[XY]=sum_xsum_yxyP(X=xwedge Y=y)$$
Note that $X$ and $Y$ are not independent here. Actually the value taken by $Y$ is completely determined by the value taken by $X$.
The term $xyP(X=xwedge Y=y)$ takes value $0$ if $x$ is even or if $y=0$
Leaving these terms out then we find:$$mathbb E[XY]=1cdot1cdot P(X=1)+3cdot1cdot P(X=3)+5cdot1cdot P(X=5)=(1+3+5)frac16=frac32$$
So that: $$mathbb E3XY=3mathbb E[XY]=frac92$$
More directly you could go for:$$mathbb E[3XY]=3mathbb E[XYmid Xtext{ is odd})P(Xtext{ is odd})+3mathbb E[XYmid Xtext{ is even})P(Xtext{ is even})=$$$$3mathbb E[Xmid Xtext{ is odd})P(Xtext{ is odd})+0=3cdotfrac13(1+3+5)frac12+0=frac92$$
answered Jan 29 at 11:32
drhabdrhab
104k545136
$endgroup$
$begingroup$
thank you so much, you fixed so many mistakes and errors i've been doing for a while in my calculations. could you please also explain to me how to calculate e[x] and e[y] separetly?
$endgroup$
– q123
Jan 29 at 11:44
$begingroup$
$mathbb EX=sum_x xP(X=x)=1cdotfrac16+2cdotfrac16+cdots+6cdotfrac16$ and $mathbb EY=0P(Y=0)+1P(Y=1)=P(Y=1)=frac12$.
$endgroup$
– drhab
Jan 29 at 12:04
add a comment |
$begingroup$
$P(X=i) = frac{1}{6}$ $P(Y=j) = frac{1}{2}$
$$E[xy]=sum_xsum_y xyP(x=i,y=j)$$
$E(3XY) = 3left(1times 1times frac{1}{6}+3times 1times frac{1}{6}+5times 1times frac{1}{6}right)$
For X = 2,4,6, Y takes the value of 0 and thus can be omitted in the calcualtion of E(XY)
$$ = 3left((1+3+5)times 1 timesfrac{1}{6}right) = frac{9}{2}$$
$$E(Y) = sum_{i=1}^{6} E(y_j=j/x_i=i)P(X=i) = (3times frac{1}{6})= frac{1}{2}$$
by the law of total expectation.
$E(Y_j/X=i) = 1$ for $X_i=1,3,5$ and $E(Y_j/X=i) = 0$ for $X_i = 2,4,6$
$$E(X) = sum_{i=1}^{6} x_iP(X=i) = frac{7}{2}$$
answered Jan 29 at 10:03
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
2
$begingroup$
But $X$ and $Y$ are not independent, so you would have to use $mathbb{P}(X = i, , Y=j)$.
$endgroup$
– Harnak
Jan 29 at 10:09
$begingroup$
If you draw the table for each x, you have a Y and that you have 6 ordered pairs and only that and in the calculation of E(XY), 3 ordered pairs become 0. and hence the result. Am I wrong?
$endgroup$
– Satish Ramanathan
Jan 29 at 10:11
$begingroup$
$mathbb P(X = 3, Y = 1) = frac16$, not $frac16 times frac12$. When $X = 3$, always $Y = 1$. Similarly for the other odd values of the die roll.
$endgroup$
– Mees de Vries
Jan 29 at 10:15
$begingroup$
You are right!!. hopefully it is correct now.
$endgroup$
– Satish Ramanathan
Jan 29 at 10:17
$begingroup$
could you please explain how to calculate e[x] and e[y] if you can?
$endgroup$
– q123
Jan 29 at 11:18
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
It is not correct to state that $mathbb E[XY]=sum_xsum_yxyP(X=x)P(Y=y)$.
It is correct to state that: $$mathbb E[XY]=sum_xsum_yxyP(X=xwedge Y=y)$$
Note that $X$ and $Y$ are not independent here. Actually the value taken by $Y$ is completely determined by the value taken by $X$.
The term $xyP(X=xwedge Y=y)$ takes value $0$ if $x$ is even or if $y=0$
Leaving these terms out then we find:$$mathbb E[XY]=1cdot1cdot P(X=1)+3cdot1cdot P(X=3)+5cdot1cdot P(X=5)=(1+3+5)frac16=frac32$$
So that: $$mathbb E3XY=3mathbb E[XY]=frac92$$
More directly you could go for:$$mathbb E[3XY]=3mathbb E[XYmid Xtext{ is odd})P(Xtext{ is odd})+3mathbb E[XYmid Xtext{ is even})P(Xtext{ is even})=$$$$3mathbb E[Xmid Xtext{ is odd})P(Xtext{ is odd})+0=3cdotfrac13(1+3+5)frac12+0=frac92$$
answered Jan 29 at 11:32
drhabdrhab
104k545136
$endgroup$
$begingroup$
thank you so much, you fixed so many mistakes and errors i've been doing for a while in my calculations. could you please also explain to me how to calculate e[x] and e[y] separetly?
$endgroup$
– q123
Jan 29 at 11:44
$begingroup$
$mathbb EX=sum_x xP(X=x)=1cdotfrac16+2cdotfrac16+cdots+6cdotfrac16$ and $mathbb EY=0P(Y=0)+1P(Y=1)=P(Y=1)=frac12$.
$endgroup$
– drhab
Jan 29 at 12:04
add a comment |
$begingroup$
It is not correct to state that $mathbb E[XY]=sum_xsum_yxyP(X=x)P(Y=y)$.
It is correct to state that: $$mathbb E[XY]=sum_xsum_yxyP(X=xwedge Y=y)$$
Note that $X$ and $Y$ are not independent here. Actually the value taken by $Y$ is completely determined by the value taken by $X$.
The term $xyP(X=xwedge Y=y)$ takes value $0$ if $x$ is even or if $y=0$
Leaving these terms out then we find:$$mathbb E[XY]=1cdot1cdot P(X=1)+3cdot1cdot P(X=3)+5cdot1cdot P(X=5)=(1+3+5)frac16=frac32$$
So that: $$mathbb E3XY=3mathbb E[XY]=frac92$$
More directly you could go for:$$mathbb E[3XY]=3mathbb E[XYmid Xtext{ is odd})P(Xtext{ is odd})+3mathbb E[XYmid Xtext{ is even})P(Xtext{ is even})=$$$$3mathbb E[Xmid Xtext{ is odd})P(Xtext{ is odd})+0=3cdotfrac13(1+3+5)frac12+0=frac92$$
answered Jan 29 at 11:32
drhabdrhab
104k545136
$endgroup$
$begingroup$
thank you so much, you fixed so many mistakes and errors i've been doing for a while in my calculations. could you please also explain to me how to calculate e[x] and e[y] separetly?
$endgroup$
– q123
Jan 29 at 11:44
$begingroup$
$mathbb EX=sum_x xP(X=x)=1cdotfrac16+2cdotfrac16+cdots+6cdotfrac16$ and $mathbb EY=0P(Y=0)+1P(Y=1)=P(Y=1)=frac12$.
$endgroup$
– drhab
Jan 29 at 12:04
add a comment |
$begingroup$
It is not correct to state that $mathbb E[XY]=sum_xsum_yxyP(X=x)P(Y=y)$.
It is correct to state that: $$mathbb E[XY]=sum_xsum_yxyP(X=xwedge Y=y)$$
Note that $X$ and $Y$ are not independent here. Actually the value taken by $Y$ is completely determined by the value taken by $X$.
The term $xyP(X=xwedge Y=y)$ takes value $0$ if $x$ is even or if $y=0$
Leaving these terms out then we find:$$mathbb E[XY]=1cdot1cdot P(X=1)+3cdot1cdot P(X=3)+5cdot1cdot P(X=5)=(1+3+5)frac16=frac32$$
So that: $$mathbb E3XY=3mathbb E[XY]=frac92$$
More directly you could go for:$$mathbb E[3XY]=3mathbb E[XYmid Xtext{ is odd})P(Xtext{ is odd})+3mathbb E[XYmid Xtext{ is even})P(Xtext{ is even})=$$$$3mathbb E[Xmid Xtext{ is odd})P(Xtext{ is odd})+0=3cdotfrac13(1+3+5)frac12+0=frac92$$
answered Jan 29 at 11:32
drhabdrhab
104k545136
$endgroup$
It is not correct to state that $mathbb E[XY]=sum_xsum_yxyP(X=x)P(Y=y)$.
It is correct to state that: $$mathbb E[XY]=sum_xsum_yxyP(X=xwedge Y=y)$$
Note that $X$ and $Y$ are not independent here. Actually the value taken by $Y$ is completely determined by the value taken by $X$.
The term $xyP(X=xwedge Y=y)$ takes value $0$ if $x$ is even or if $y=0$
Leaving these terms out then we find:$$mathbb E[XY]=1cdot1cdot P(X=1)+3cdot1cdot P(X=3)+5cdot1cdot P(X=5)=(1+3+5)frac16=frac32$$
So that: $$mathbb E3XY=3mathbb E[XY]=frac92$$
More directly you could go for:$$mathbb E[3XY]=3mathbb E[XYmid Xtext{ is odd})P(Xtext{ is odd})+3mathbb E[XYmid Xtext{ is even})P(Xtext{ is even})=$$$$3mathbb E[Xmid Xtext{ is odd})P(Xtext{ is odd})+0=3cdotfrac13(1+3+5)frac12+0=frac92$$
answered Jan 29 at 11:32
drhabdrhab
104k545136
edited Jan 29 at 14:31
answered Jan 29 at 11:32
drhabdrhab
104k545136
answered Jan 29 at 11:32
drhabdrhab
104k545136
answered Jan 29 at 11:32
drhabdrhab
104k545136
104k545136
$begingroup$
thank you so much, you fixed so many mistakes and errors i've been doing for a while in my calculations. could you please also explain to me how to calculate e[x] and e[y] separetly?
$endgroup$
– q123
Jan 29 at 11:44
$begingroup$
$mathbb EX=sum_x xP(X=x)=1cdotfrac16+2cdotfrac16+cdots+6cdotfrac16$ and $mathbb EY=0P(Y=0)+1P(Y=1)=P(Y=1)=frac12$.
$endgroup$
– drhab
Jan 29 at 12:04
add a comment |
$begingroup$
thank you so much, you fixed so many mistakes and errors i've been doing for a while in my calculations. could you please also explain to me how to calculate e[x] and e[y] separetly?
$endgroup$
– q123
Jan 29 at 11:44
$begingroup$
$mathbb EX=sum_x xP(X=x)=1cdotfrac16+2cdotfrac16+cdots+6cdotfrac16$ and $mathbb EY=0P(Y=0)+1P(Y=1)=P(Y=1)=frac12$.
$endgroup$
– drhab
Jan 29 at 12:04
$begingroup$
thank you so much, you fixed so many mistakes and errors i've been doing for a while in my calculations. could you please also explain to me how to calculate e[x] and e[y] separetly?
$endgroup$
– q123
Jan 29 at 11:44
$begingroup$
thank you so much, you fixed so many mistakes and errors i've been doing for a while in my calculations. could you please also explain to me how to calculate e[x] and e[y] separetly?
$endgroup$
– q123
Jan 29 at 11:44
$begingroup$
$mathbb EX=sum_x xP(X=x)=1cdotfrac16+2cdotfrac16+cdots+6cdotfrac16$ and $mathbb EY=0P(Y=0)+1P(Y=1)=P(Y=1)=frac12$.
$endgroup$
– drhab
Jan 29 at 12:04
$begingroup$
$mathbb EX=sum_x xP(X=x)=1cdotfrac16+2cdotfrac16+cdots+6cdotfrac16$ and $mathbb EY=0P(Y=0)+1P(Y=1)=P(Y=1)=frac12$.
$endgroup$
– drhab
Jan 29 at 12:04
add a comment |
$begingroup$
$P(X=i) = frac{1}{6}$ $P(Y=j) = frac{1}{2}$
$$E[xy]=sum_xsum_y xyP(x=i,y=j)$$
$E(3XY) = 3left(1times 1times frac{1}{6}+3times 1times frac{1}{6}+5times 1times frac{1}{6}right)$
For X = 2,4,6, Y takes the value of 0 and thus can be omitted in the calcualtion of E(XY)
$$ = 3left((1+3+5)times 1 timesfrac{1}{6}right) = frac{9}{2}$$
$$E(Y) = sum_{i=1}^{6} E(y_j=j/x_i=i)P(X=i) = (3times frac{1}{6})= frac{1}{2}$$
by the law of total expectation.
$E(Y_j/X=i) = 1$ for $X_i=1,3,5$ and $E(Y_j/X=i) = 0$ for $X_i = 2,4,6$
$$E(X) = sum_{i=1}^{6} x_iP(X=i) = frac{7}{2}$$
answered Jan 29 at 10:03
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
2
$begingroup$
But $X$ and $Y$ are not independent, so you would have to use $mathbb{P}(X = i, , Y=j)$.
$endgroup$
– Harnak
Jan 29 at 10:09
$begingroup$
If you draw the table for each x, you have a Y and that you have 6 ordered pairs and only that and in the calculation of E(XY), 3 ordered pairs become 0. and hence the result. Am I wrong?
$endgroup$
– Satish Ramanathan
Jan 29 at 10:11
$begingroup$
$mathbb P(X = 3, Y = 1) = frac16$, not $frac16 times frac12$. When $X = 3$, always $Y = 1$. Similarly for the other odd values of the die roll.
$endgroup$
– Mees de Vries
Jan 29 at 10:15
$begingroup$
You are right!!. hopefully it is correct now.
$endgroup$
– Satish Ramanathan
Jan 29 at 10:17
$begingroup$
could you please explain how to calculate e[x] and e[y] if you can?
$endgroup$
– q123
Jan 29 at 11:18
add a comment |
$begingroup$
$P(X=i) = frac{1}{6}$ $P(Y=j) = frac{1}{2}$
$$E[xy]=sum_xsum_y xyP(x=i,y=j)$$
$E(3XY) = 3left(1times 1times frac{1}{6}+3times 1times frac{1}{6}+5times 1times frac{1}{6}right)$
For X = 2,4,6, Y takes the value of 0 and thus can be omitted in the calcualtion of E(XY)
$$ = 3left((1+3+5)times 1 timesfrac{1}{6}right) = frac{9}{2}$$
$$E(Y) = sum_{i=1}^{6} E(y_j=j/x_i=i)P(X=i) = (3times frac{1}{6})= frac{1}{2}$$
by the law of total expectation.
$E(Y_j/X=i) = 1$ for $X_i=1,3,5$ and $E(Y_j/X=i) = 0$ for $X_i = 2,4,6$
$$E(X) = sum_{i=1}^{6} x_iP(X=i) = frac{7}{2}$$
answered Jan 29 at 10:03
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
2
$begingroup$
But $X$ and $Y$ are not independent, so you would have to use $mathbb{P}(X = i, , Y=j)$.
$endgroup$
– Harnak
Jan 29 at 10:09
$begingroup$
If you draw the table for each x, you have a Y and that you have 6 ordered pairs and only that and in the calculation of E(XY), 3 ordered pairs become 0. and hence the result. Am I wrong?
$endgroup$
– Satish Ramanathan
Jan 29 at 10:11
$begingroup$
$mathbb P(X = 3, Y = 1) = frac16$, not $frac16 times frac12$. When $X = 3$, always $Y = 1$. Similarly for the other odd values of the die roll.
$endgroup$
– Mees de Vries
Jan 29 at 10:15
$begingroup$
You are right!!. hopefully it is correct now.
$endgroup$
– Satish Ramanathan
Jan 29 at 10:17
$begingroup$
could you please explain how to calculate e[x] and e[y] if you can?
$endgroup$
– q123
Jan 29 at 11:18
add a comment |
$begingroup$
$P(X=i) = frac{1}{6}$ $P(Y=j) = frac{1}{2}$
$$E[xy]=sum_xsum_y xyP(x=i,y=j)$$
$E(3XY) = 3left(1times 1times frac{1}{6}+3times 1times frac{1}{6}+5times 1times frac{1}{6}right)$
For X = 2,4,6, Y takes the value of 0 and thus can be omitted in the calcualtion of E(XY)
$$ = 3left((1+3+5)times 1 timesfrac{1}{6}right) = frac{9}{2}$$
$$E(Y) = sum_{i=1}^{6} E(y_j=j/x_i=i)P(X=i) = (3times frac{1}{6})= frac{1}{2}$$
by the law of total expectation.
$E(Y_j/X=i) = 1$ for $X_i=1,3,5$ and $E(Y_j/X=i) = 0$ for $X_i = 2,4,6$
$$E(X) = sum_{i=1}^{6} x_iP(X=i) = frac{7}{2}$$
answered Jan 29 at 10:03
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
$P(X=i) = frac{1}{6}$ $P(Y=j) = frac{1}{2}$
$$E[xy]=sum_xsum_y xyP(x=i,y=j)$$
$E(3XY) = 3left(1times 1times frac{1}{6}+3times 1times frac{1}{6}+5times 1times frac{1}{6}right)$
For X = 2,4,6, Y takes the value of 0 and thus can be omitted in the calcualtion of E(XY)
$$ = 3left((1+3+5)times 1 timesfrac{1}{6}right) = frac{9}{2}$$
$$E(Y) = sum_{i=1}^{6} E(y_j=j/x_i=i)P(X=i) = (3times frac{1}{6})= frac{1}{2}$$
by the law of total expectation.
$E(Y_j/X=i) = 1$ for $X_i=1,3,5$ and $E(Y_j/X=i) = 0$ for $X_i = 2,4,6$
$$E(X) = sum_{i=1}^{6} x_iP(X=i) = frac{7}{2}$$
answered Jan 29 at 10:03
Satish RamanathanSatish Ramanathan
10k31323
edited Jan 29 at 11:32
answered Jan 29 at 10:03
Satish RamanathanSatish Ramanathan
10k31323
answered Jan 29 at 10:03
Satish RamanathanSatish Ramanathan
10k31323
answered Jan 29 at 10:03
Satish RamanathanSatish Ramanathan
10k31323
10k31323
2
$begingroup$
But $X$ and $Y$ are not independent, so you would have to use $mathbb{P}(X = i, , Y=j)$.
$endgroup$
– Harnak
Jan 29 at 10:09
$begingroup$
If you draw the table for each x, you have a Y and that you have 6 ordered pairs and only that and in the calculation of E(XY), 3 ordered pairs become 0. and hence the result. Am I wrong?
$endgroup$
– Satish Ramanathan
Jan 29 at 10:11
$begingroup$
$mathbb P(X = 3, Y = 1) = frac16$, not $frac16 times frac12$. When $X = 3$, always $Y = 1$. Similarly for the other odd values of the die roll.
$endgroup$
– Mees de Vries
Jan 29 at 10:15
$begingroup$
You are right!!. hopefully it is correct now.
$endgroup$
– Satish Ramanathan
Jan 29 at 10:17
$begingroup$
could you please explain how to calculate e[x] and e[y] if you can?
$endgroup$
– q123
Jan 29 at 11:18
add a comment |
2
$begingroup$
But $X$ and $Y$ are not independent, so you would have to use $mathbb{P}(X = i, , Y=j)$.
$endgroup$
– Harnak
Jan 29 at 10:09
$begingroup$
If you draw the table for each x, you have a Y and that you have 6 ordered pairs and only that and in the calculation of E(XY), 3 ordered pairs become 0. and hence the result. Am I wrong?
$endgroup$
– Satish Ramanathan
Jan 29 at 10:11
$begingroup$
$mathbb P(X = 3, Y = 1) = frac16$, not $frac16 times frac12$. When $X = 3$, always $Y = 1$. Similarly for the other odd values of the die roll.
$endgroup$
– Mees de Vries
Jan 29 at 10:15
$begingroup$
You are right!!. hopefully it is correct now.
$endgroup$
– Satish Ramanathan
Jan 29 at 10:17
$begingroup$
could you please explain how to calculate e[x] and e[y] if you can?
$endgroup$
– q123
Jan 29 at 11:18
2
2
$begingroup$
But $X$ and $Y$ are not independent, so you would have to use $mathbb{P}(X = i, , Y=j)$.
$endgroup$
– Harnak
Jan 29 at 10:09
$begingroup$
But $X$ and $Y$ are not independent, so you would have to use $mathbb{P}(X = i, , Y=j)$.
$endgroup$
– Harnak
Jan 29 at 10:09
$begingroup$
If you draw the table for each x, you have a Y and that you have 6 ordered pairs and only that and in the calculation of E(XY), 3 ordered pairs become 0. and hence the result. Am I wrong?
$endgroup$
– Satish Ramanathan
Jan 29 at 10:11
$begingroup$
If you draw the table for each x, you have a Y and that you have 6 ordered pairs and only that and in the calculation of E(XY), 3 ordered pairs become 0. and hence the result. Am I wrong?
$endgroup$
– Satish Ramanathan
Jan 29 at 10:11
$begingroup$
$mathbb P(X = 3, Y = 1) = frac16$, not $frac16 times frac12$. When $X = 3$, always $Y = 1$. Similarly for the other odd values of the die roll.
$endgroup$
– Mees de Vries
Jan 29 at 10:15
$begingroup$
$mathbb P(X = 3, Y = 1) = frac16$, not $frac16 times frac12$. When $X = 3$, always $Y = 1$. Similarly for the other odd values of the die roll.
$endgroup$
– Mees de Vries
Jan 29 at 10:15
$begingroup$
You are right!!. hopefully it is correct now.
$endgroup$
– Satish Ramanathan
Jan 29 at 10:17
$begingroup$
You are right!!. hopefully it is correct now.
$endgroup$
– Satish Ramanathan
Jan 29 at 10:17
$begingroup$
could you please explain how to calculate e[x] and e[y] if you can?
$endgroup$
– q123
Jan 29 at 11:18
$begingroup$
could you please explain how to calculate e[x] and e[y] if you can?
$endgroup$
– q123
Jan 29 at 11:18
add a comment |
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$begingroup$
@SatishRamanathan edited. does it make more sense now?
$endgroup$
– q123
Jan 29 at 9:28
$begingroup$
I don't really get what the values for $X$ and $Y$ are. What value does $X$ have for an odd result? Same question for $Y$ on an even result.
$endgroup$
– Harnak
Jan 29 at 9:40
$begingroup$
bah, i was wonering that the same until i looked at the errata. now it makes more sense. editing again
$endgroup$
– q123
Jan 29 at 9:43
$begingroup$
i got that $e[x] = 16*(1+2+...+6) = 3.5$ and $e[y]= 1*(1+3+5)*1/6 + 0*16*(2+4+6) = 9/6=1.5$. i think that e[y] should be 0.5, because of the table, but i don't know how to show it mathmatically
$endgroup$
– q123
Jan 29 at 10:00