$E[E[Xmid Y] =$ $sum_ymid P(Y=y>0)E[Xmid Y=y] cdot P(Y =y)$

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I'm looking at the proof of

$$E[X] = E[E[Xmid Y]]$$

But I'm having trouble to get why does, for example if we take X,Y discrete random variables, we have that

$$E[E[Xmid Y]]= sum_ymid P(Y=y)>0E[Xmid Y=y] cdot P(Y =y)$$

I know that

$E[Xmid Y]$ can be defined as a random variable $E[Xmid Y=y](omega)$ if $ Y(omega) = y$

but from that I lose a bit of intuition.

Does it mean that if we want the expected value of $E[Xmid Y]$ knowing that its a random variable where the variable is $y in operatornameIm(Y)$ then, we must simply all the "probability space of the expected value of $X$ knowing $Y = y$?

Basically, if someone has a good intuitive explanation I'd be so happy!

Also, it is written that

$$sum_ymid P(Y=y)>0E[Xmid Y=y]cdot P(Y =y) = sum_ymid P(Y=y)>0 fracE[Xcdot mathbb1_(Y=y)]P(Y=y) P(Y=y)$$

(where $1$ is the indicating function)

$$= sum_ymid P(Y=y)>0E[Xmidmathbb1_(Y=y)] = E[Xmidmathbb1_(Y=y)] = E[X]$$

and i'm also having trouble following that part.

Thank you all for everything!

probability probability-theory proof-writing

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edited Jul 18 at 23:51

Michael Hardy

204k23186462

asked Jul 18 at 21:23

Ian Leclaire

1489

add a comment | 

up vote
1
down vote

favorite

I'm looking at the proof of

$$E[X] = E[E[Xmid Y]]$$

But I'm having trouble to get why does, for example if we take X,Y discrete random variables, we have that

$$E[E[Xmid Y]]= sum_ymid P(Y=y)>0E[Xmid Y=y] cdot P(Y =y)$$

I know that

$E[Xmid Y]$ can be defined as a random variable $E[Xmid Y=y](omega)$ if $ Y(omega) = y$

but from that I lose a bit of intuition.

Does it mean that if we want the expected value of $E[Xmid Y]$ knowing that its a random variable where the variable is $y in operatornameIm(Y)$ then, we must simply all the "probability space of the expected value of $X$ knowing $Y = y$?

Basically, if someone has a good intuitive explanation I'd be so happy!

Also, it is written that

$$sum_ymid P(Y=y)>0E[Xmid Y=y]cdot P(Y =y) = sum_ymid P(Y=y)>0 fracE[Xcdot mathbb1_(Y=y)]P(Y=y) P(Y=y)$$

(where $1$ is the indicating function)

$$= sum_ymid P(Y=y)>0E[Xmidmathbb1_(Y=y)] = E[Xmidmathbb1_(Y=y)] = E[X]$$

and i'm also having trouble following that part.

Thank you all for everything!

probability probability-theory proof-writing

share|cite|improve this question

edited Jul 18 at 23:51

Michael Hardy

204k23186462

asked Jul 18 at 21:23

Ian Leclaire

1489

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I'm looking at the proof of

$$E[X] = E[E[Xmid Y]]$$

But I'm having trouble to get why does, for example if we take X,Y discrete random variables, we have that

$$E[E[Xmid Y]]= sum_ymid P(Y=y)>0E[Xmid Y=y] cdot P(Y =y)$$

I know that

$E[Xmid Y]$ can be defined as a random variable $E[Xmid Y=y](omega)$ if $ Y(omega) = y$

but from that I lose a bit of intuition.

Does it mean that if we want the expected value of $E[Xmid Y]$ knowing that its a random variable where the variable is $y in operatornameIm(Y)$ then, we must simply all the "probability space of the expected value of $X$ knowing $Y = y$?

Basically, if someone has a good intuitive explanation I'd be so happy!

Also, it is written that

$$sum_ymid P(Y=y)>0E[Xmid Y=y]cdot P(Y =y) = sum_ymid P(Y=y)>0 fracE[Xcdot mathbb1_(Y=y)]P(Y=y) P(Y=y)$$

(where $1$ is the indicating function)

$$= sum_ymid P(Y=y)>0E[Xmidmathbb1_(Y=y)] = E[Xmidmathbb1_(Y=y)] = E[X]$$

and i'm also having trouble following that part.

Thank you all for everything!

probability probability-theory proof-writing

share|cite|improve this question

edited Jul 18 at 23:51

Michael Hardy

204k23186462

asked Jul 18 at 21:23

Ian Leclaire

1489

I'm looking at the proof of

$$E[X] = E[E[Xmid Y]]$$

But I'm having trouble to get why does, for example if we take X,Y discrete random variables, we have that

$$E[E[Xmid Y]]= sum_ymid P(Y=y)>0E[Xmid Y=y] cdot P(Y =y)$$

I know that

$E[Xmid Y]$ can be defined as a random variable $E[Xmid Y=y](omega)$ if $ Y(omega) = y$

but from that I lose a bit of intuition.

Does it mean that if we want the expected value of $E[Xmid Y]$ knowing that its a random variable where the variable is $y in operatornameIm(Y)$ then, we must simply all the "probability space of the expected value of $X$ knowing $Y = y$?

Basically, if someone has a good intuitive explanation I'd be so happy!

Also, it is written that

$$sum_ymid P(Y=y)>0E[Xmid Y=y]cdot P(Y =y) = sum_ymid P(Y=y)>0 fracE[Xcdot mathbb1_(Y=y)]P(Y=y) P(Y=y)$$

(where $1$ is the indicating function)

$$= sum_ymid P(Y=y)>0E[Xmidmathbb1_(Y=y)] = E[Xmidmathbb1_(Y=y)] = E[X]$$

and i'm also having trouble following that part.

Thank you all for everything!

probability probability-theory proof-writing

share|cite|improve this question

edited Jul 18 at 23:51

Michael Hardy

204k23186462

asked Jul 18 at 21:23

Ian Leclaire

1489

share|cite|improve this question

share|cite|improve this question

edited Jul 18 at 23:51

Michael Hardy

204k23186462

edited Jul 18 at 23:51

Michael Hardy

204k23186462

edited Jul 18 at 23:51

Michael Hardy

204k23186462

204k23186462

asked Jul 18 at 21:23

Ian Leclaire

1489

asked Jul 18 at 21:23

Ian Leclaire

1489

asked Jul 18 at 21:23

Ian Leclaire

1489

1489

add a comment | 

add a comment | 

2 Answers
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oldest

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accepted

For the first equation:

$$E[E[Xmid Y]]= sum_ymid P(Y=y)>0E[Xmid Y=y] cdot P(Y =y)$$

It helps to think of $E[Xmid Y]$ as a function $f(Y)$, which is a random variable (see here). Then, as for all expected values, you iterate over all the possible values of the random variable multiplying its probability:

$$E[f(Y)]= sum_y mid P(Y=y)>0 f(y) cdot P(Y =y)$$

For your doubt on your last equations, I will write the proof in another way. Maybe it helps you.

$$E[E[Xmid Y]] = sum_y mid P(Y=y)>0 E[Xmid Y = y] cdot P(Y = y) = sum_y mid P(Y=y)>0 ( sum_x in X x P(X = x mid Y = y))cdot P(Y = y)$$

$$= sum_ymid P(Y=y)>0 (sum_x in X x dfracP(X = x; Y = y)P(Y = y))cdot P(Y = y) = sum_x in X x sum_y mid P(Y=y)>0 P(X = x; Y = y)$$

$$ = sum_x in X x P(X = x) = E[X]$$

share|cite|improve this answer

edited Jul 19 at 2:38

Did

242k23208443

answered Jul 18 at 23:11

Enzo Nakamura

605

add a comment | 

up vote
1
down vote

I know that $E[Xmid Y]$ can be defined as a random variable $E[Xmid Y=y](omega)$ if $ Y(omega) = y$

Not quite; it is the other way.   $mathsf E[Xmid Y=y]$ is defined as the value of random variable, $mathsf E[Xmid Y],$ for all outcomes, $omega$, where $Y(omega)=y$. $forall omegain Y^-1(y): mathsf E(Xmid Y)(omega)=mathsf E(Xmid Y=y)$

So $$beginalignmathsf E(mathsf E(Xmid Y))
&= sum_omegainOmega mathsf E(Xmid Y)(omega)cdotmathsf Pomega &&textby definition
\[1ex]&= sum_yin Y(Omega)sum_omegain Y^-1(y) mathsf E(Xmid Y)(omega)cdotmathsf Pomega &&textPartitioning the series
\[1ex] &= sum_yin Y(Omega) mathsf E(Xmid Y=y) sum_omegain Y^-1(Omega)mathsf Pomega&&textby definition of mathsf E(Xmid Y=y)
\[1ex] &=sum_yin Y(Omega)mathsf E(Xmid Y=y)cdotmathsf PomegainOmega:Y(omega)=y &&textby countable additivity
\[1ex] &=sum_yin Y(Omega)mathsf E(Xmid Y=y)cdotmathsf P(Y=y) &&textAbreviationendalign$$

Now, what kind of value is $mathsf E(Xmid Y=y)$?   Well for any event $E$ with nonzero probability measure, we define $mathsf E(Xmid E)=mathsf E(Xmathbf 1_E)divmathsf P(E)$.

$$beginalignmathsf E(mathsf E(Xmid Y))&=sum_y:mathsf P(Y=y)>0 dfracmathsf E(Xmathbf 1_Y=y)mathsf P(Y=y)mathsf P(Y=y)+sum_y:mathsf P(Y=y)=00
\[1ex] &= sum_y:mathsf P(Y=y)>0 mathsf E(Xmathbf 1_Y=y)
\[1ex] &= sum_y:mathsf P(Y=y)>0sum_omegainOmega:Y(omega)=y X(omega)mathsf Pomega
\[1ex] &= sum_omegainOmegaX(omega)mathsf Pomega
\[1ex] &= mathsf E(X)endalign$$

share|cite|improve this answer

answered Jul 19 at 2:15

Graham Kemp

80.1k43275

add a comment | 

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2 Answers
2

active

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votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

For the first equation:

$$E[E[Xmid Y]]= sum_ymid P(Y=y)>0E[Xmid Y=y] cdot P(Y =y)$$

It helps to think of $E[Xmid Y]$ as a function $f(Y)$, which is a random variable (see here). Then, as for all expected values, you iterate over all the possible values of the random variable multiplying its probability:

$$E[f(Y)]= sum_y mid P(Y=y)>0 f(y) cdot P(Y =y)$$

For your doubt on your last equations, I will write the proof in another way. Maybe it helps you.

$$E[E[Xmid Y]] = sum_y mid P(Y=y)>0 E[Xmid Y = y] cdot P(Y = y) = sum_y mid P(Y=y)>0 ( sum_x in X x P(X = x mid Y = y))cdot P(Y = y)$$

$$= sum_ymid P(Y=y)>0 (sum_x in X x dfracP(X = x; Y = y)P(Y = y))cdot P(Y = y) = sum_x in X x sum_y mid P(Y=y)>0 P(X = x; Y = y)$$

$$ = sum_x in X x P(X = x) = E[X]$$

share|cite|improve this answer

edited Jul 19 at 2:38

Did

242k23208443

answered Jul 18 at 23:11

Enzo Nakamura

605

add a comment | 

up vote
2
down vote



accepted

For the first equation:

$$E[E[Xmid Y]]= sum_ymid P(Y=y)>0E[Xmid Y=y] cdot P(Y =y)$$

It helps to think of $E[Xmid Y]$ as a function $f(Y)$, which is a random variable (see here). Then, as for all expected values, you iterate over all the possible values of the random variable multiplying its probability:

$$E[f(Y)]= sum_y mid P(Y=y)>0 f(y) cdot P(Y =y)$$

For your doubt on your last equations, I will write the proof in another way. Maybe it helps you.

$$E[E[Xmid Y]] = sum_y mid P(Y=y)>0 E[Xmid Y = y] cdot P(Y = y) = sum_y mid P(Y=y)>0 ( sum_x in X x P(X = x mid Y = y))cdot P(Y = y)$$

$$= sum_ymid P(Y=y)>0 (sum_x in X x dfracP(X = x; Y = y)P(Y = y))cdot P(Y = y) = sum_x in X x sum_y mid P(Y=y)>0 P(X = x; Y = y)$$

$$ = sum_x in X x P(X = x) = E[X]$$

share|cite|improve this answer

edited Jul 19 at 2:38

Did

242k23208443

answered Jul 18 at 23:11

Enzo Nakamura

605

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

For the first equation:

$$E[E[Xmid Y]]= sum_ymid P(Y=y)>0E[Xmid Y=y] cdot P(Y =y)$$

It helps to think of $E[Xmid Y]$ as a function $f(Y)$, which is a random variable (see here). Then, as for all expected values, you iterate over all the possible values of the random variable multiplying its probability:

$$E[f(Y)]= sum_y mid P(Y=y)>0 f(y) cdot P(Y =y)$$

For your doubt on your last equations, I will write the proof in another way. Maybe it helps you.

$$E[E[Xmid Y]] = sum_y mid P(Y=y)>0 E[Xmid Y = y] cdot P(Y = y) = sum_y mid P(Y=y)>0 ( sum_x in X x P(X = x mid Y = y))cdot P(Y = y)$$

$$= sum_ymid P(Y=y)>0 (sum_x in X x dfracP(X = x; Y = y)P(Y = y))cdot P(Y = y) = sum_x in X x sum_y mid P(Y=y)>0 P(X = x; Y = y)$$

$$ = sum_x in X x P(X = x) = E[X]$$

share|cite|improve this answer

edited Jul 19 at 2:38

Did

242k23208443

answered Jul 18 at 23:11

Enzo Nakamura

605

For the first equation:

$$E[E[Xmid Y]]= sum_ymid P(Y=y)>0E[Xmid Y=y] cdot P(Y =y)$$

It helps to think of $E[Xmid Y]$ as a function $f(Y)$, which is a random variable (see here). Then, as for all expected values, you iterate over all the possible values of the random variable multiplying its probability:

$$E[f(Y)]= sum_y mid P(Y=y)>0 f(y) cdot P(Y =y)$$

For your doubt on your last equations, I will write the proof in another way. Maybe it helps you.

$$E[E[Xmid Y]] = sum_y mid P(Y=y)>0 E[Xmid Y = y] cdot P(Y = y) = sum_y mid P(Y=y)>0 ( sum_x in X x P(X = x mid Y = y))cdot P(Y = y)$$

$$= sum_ymid P(Y=y)>0 (sum_x in X x dfracP(X = x; Y = y)P(Y = y))cdot P(Y = y) = sum_x in X x sum_y mid P(Y=y)>0 P(X = x; Y = y)$$

$$ = sum_x in X x P(X = x) = E[X]$$

share|cite|improve this answer

edited Jul 19 at 2:38

Did

242k23208443

answered Jul 18 at 23:11

Enzo Nakamura

605

share|cite|improve this answer

share|cite|improve this answer

edited Jul 19 at 2:38

Did

242k23208443

edited Jul 19 at 2:38

Did

242k23208443

edited Jul 19 at 2:38

Did

242k23208443

242k23208443

answered Jul 18 at 23:11

Enzo Nakamura

605

answered Jul 18 at 23:11

Enzo Nakamura

605

answered Jul 18 at 23:11

Enzo Nakamura

605

605

add a comment | 

add a comment | 

up vote
1
down vote

I know that $E[Xmid Y]$ can be defined as a random variable $E[Xmid Y=y](omega)$ if $ Y(omega) = y$

Not quite; it is the other way.   $mathsf E[Xmid Y=y]$ is defined as the value of random variable, $mathsf E[Xmid Y],$ for all outcomes, $omega$, where $Y(omega)=y$. $forall omegain Y^-1(y): mathsf E(Xmid Y)(omega)=mathsf E(Xmid Y=y)$

So $$beginalignmathsf E(mathsf E(Xmid Y))
&= sum_omegainOmega mathsf E(Xmid Y)(omega)cdotmathsf Pomega &&textby definition
\[1ex]&= sum_yin Y(Omega)sum_omegain Y^-1(y) mathsf E(Xmid Y)(omega)cdotmathsf Pomega &&textPartitioning the series
\[1ex] &= sum_yin Y(Omega) mathsf E(Xmid Y=y) sum_omegain Y^-1(Omega)mathsf Pomega&&textby definition of mathsf E(Xmid Y=y)
\[1ex] &=sum_yin Y(Omega)mathsf E(Xmid Y=y)cdotmathsf PomegainOmega:Y(omega)=y &&textby countable additivity
\[1ex] &=sum_yin Y(Omega)mathsf E(Xmid Y=y)cdotmathsf P(Y=y) &&textAbreviationendalign$$

Now, what kind of value is $mathsf E(Xmid Y=y)$?   Well for any event $E$ with nonzero probability measure, we define $mathsf E(Xmid E)=mathsf E(Xmathbf 1_E)divmathsf P(E)$.

$$beginalignmathsf E(mathsf E(Xmid Y))&=sum_y:mathsf P(Y=y)>0 dfracmathsf E(Xmathbf 1_Y=y)mathsf P(Y=y)mathsf P(Y=y)+sum_y:mathsf P(Y=y)=00
\[1ex] &= sum_y:mathsf P(Y=y)>0 mathsf E(Xmathbf 1_Y=y)
\[1ex] &= sum_y:mathsf P(Y=y)>0sum_omegainOmega:Y(omega)=y X(omega)mathsf Pomega
\[1ex] &= sum_omegainOmegaX(omega)mathsf Pomega
\[1ex] &= mathsf E(X)endalign$$

share|cite|improve this answer

answered Jul 19 at 2:15

Graham Kemp

80.1k43275

add a comment | 

up vote
1
down vote

I know that $E[Xmid Y]$ can be defined as a random variable $E[Xmid Y=y](omega)$ if $ Y(omega) = y$

Not quite; it is the other way.   $mathsf E[Xmid Y=y]$ is defined as the value of random variable, $mathsf E[Xmid Y],$ for all outcomes, $omega$, where $Y(omega)=y$. $forall omegain Y^-1(y): mathsf E(Xmid Y)(omega)=mathsf E(Xmid Y=y)$

So $$beginalignmathsf E(mathsf E(Xmid Y))
&= sum_omegainOmega mathsf E(Xmid Y)(omega)cdotmathsf Pomega &&textby definition
\[1ex]&= sum_yin Y(Omega)sum_omegain Y^-1(y) mathsf E(Xmid Y)(omega)cdotmathsf Pomega &&textPartitioning the series
\[1ex] &= sum_yin Y(Omega) mathsf E(Xmid Y=y) sum_omegain Y^-1(Omega)mathsf Pomega&&textby definition of mathsf E(Xmid Y=y)
\[1ex] &=sum_yin Y(Omega)mathsf E(Xmid Y=y)cdotmathsf PomegainOmega:Y(omega)=y &&textby countable additivity
\[1ex] &=sum_yin Y(Omega)mathsf E(Xmid Y=y)cdotmathsf P(Y=y) &&textAbreviationendalign$$

Now, what kind of value is $mathsf E(Xmid Y=y)$?   Well for any event $E$ with nonzero probability measure, we define $mathsf E(Xmid E)=mathsf E(Xmathbf 1_E)divmathsf P(E)$.

$$beginalignmathsf E(mathsf E(Xmid Y))&=sum_y:mathsf P(Y=y)>0 dfracmathsf E(Xmathbf 1_Y=y)mathsf P(Y=y)mathsf P(Y=y)+sum_y:mathsf P(Y=y)=00
\[1ex] &= sum_y:mathsf P(Y=y)>0 mathsf E(Xmathbf 1_Y=y)
\[1ex] &= sum_y:mathsf P(Y=y)>0sum_omegainOmega:Y(omega)=y X(omega)mathsf Pomega
\[1ex] &= sum_omegainOmegaX(omega)mathsf Pomega
\[1ex] &= mathsf E(X)endalign$$

share|cite|improve this answer

answered Jul 19 at 2:15

Graham Kemp

80.1k43275

add a comment | 

up vote
1
down vote

up vote
1
down vote

I know that $E[Xmid Y]$ can be defined as a random variable $E[Xmid Y=y](omega)$ if $ Y(omega) = y$

Not quite; it is the other way.   $mathsf E[Xmid Y=y]$ is defined as the value of random variable, $mathsf E[Xmid Y],$ for all outcomes, $omega$, where $Y(omega)=y$. $forall omegain Y^-1(y): mathsf E(Xmid Y)(omega)=mathsf E(Xmid Y=y)$

So $$beginalignmathsf E(mathsf E(Xmid Y))
&= sum_omegainOmega mathsf E(Xmid Y)(omega)cdotmathsf Pomega &&textby definition
\[1ex]&= sum_yin Y(Omega)sum_omegain Y^-1(y) mathsf E(Xmid Y)(omega)cdotmathsf Pomega &&textPartitioning the series
\[1ex] &= sum_yin Y(Omega) mathsf E(Xmid Y=y) sum_omegain Y^-1(Omega)mathsf Pomega&&textby definition of mathsf E(Xmid Y=y)
\[1ex] &=sum_yin Y(Omega)mathsf E(Xmid Y=y)cdotmathsf PomegainOmega:Y(omega)=y &&textby countable additivity
\[1ex] &=sum_yin Y(Omega)mathsf E(Xmid Y=y)cdotmathsf P(Y=y) &&textAbreviationendalign$$

Now, what kind of value is $mathsf E(Xmid Y=y)$?   Well for any event $E$ with nonzero probability measure, we define $mathsf E(Xmid E)=mathsf E(Xmathbf 1_E)divmathsf P(E)$.

$$beginalignmathsf E(mathsf E(Xmid Y))&=sum_y:mathsf P(Y=y)>0 dfracmathsf E(Xmathbf 1_Y=y)mathsf P(Y=y)mathsf P(Y=y)+sum_y:mathsf P(Y=y)=00
\[1ex] &= sum_y:mathsf P(Y=y)>0 mathsf E(Xmathbf 1_Y=y)
\[1ex] &= sum_y:mathsf P(Y=y)>0sum_omegainOmega:Y(omega)=y X(omega)mathsf Pomega
\[1ex] &= sum_omegainOmegaX(omega)mathsf Pomega
\[1ex] &= mathsf E(X)endalign$$

share|cite|improve this answer

answered Jul 19 at 2:15

Graham Kemp

80.1k43275

I know that $E[Xmid Y]$ can be defined as a random variable $E[Xmid Y=y](omega)$ if $ Y(omega) = y$

Not quite; it is the other way.   $mathsf E[Xmid Y=y]$ is defined as the value of random variable, $mathsf E[Xmid Y],$ for all outcomes, $omega$, where $Y(omega)=y$. $forall omegain Y^-1(y): mathsf E(Xmid Y)(omega)=mathsf E(Xmid Y=y)$

So $$beginalignmathsf E(mathsf E(Xmid Y))
&= sum_omegainOmega mathsf E(Xmid Y)(omega)cdotmathsf Pomega &&textby definition
\[1ex]&= sum_yin Y(Omega)sum_omegain Y^-1(y) mathsf E(Xmid Y)(omega)cdotmathsf Pomega &&textPartitioning the series
\[1ex] &= sum_yin Y(Omega) mathsf E(Xmid Y=y) sum_omegain Y^-1(Omega)mathsf Pomega&&textby definition of mathsf E(Xmid Y=y)
\[1ex] &=sum_yin Y(Omega)mathsf E(Xmid Y=y)cdotmathsf PomegainOmega:Y(omega)=y &&textby countable additivity
\[1ex] &=sum_yin Y(Omega)mathsf E(Xmid Y=y)cdotmathsf P(Y=y) &&textAbreviationendalign$$

Now, what kind of value is $mathsf E(Xmid Y=y)$?   Well for any event $E$ with nonzero probability measure, we define $mathsf E(Xmid E)=mathsf E(Xmathbf 1_E)divmathsf P(E)$.

$$beginalignmathsf E(mathsf E(Xmid Y))&=sum_y:mathsf P(Y=y)>0 dfracmathsf E(Xmathbf 1_Y=y)mathsf P(Y=y)mathsf P(Y=y)+sum_y:mathsf P(Y=y)=00
\[1ex] &= sum_y:mathsf P(Y=y)>0 mathsf E(Xmathbf 1_Y=y)
\[1ex] &= sum_y:mathsf P(Y=y)>0sum_omegainOmega:Y(omega)=y X(omega)mathsf Pomega
\[1ex] &= sum_omegainOmegaX(omega)mathsf Pomega
\[1ex] &= mathsf E(X)endalign$$

share|cite|improve this answer

answered Jul 19 at 2:15

Graham Kemp

80.1k43275

share|cite|improve this answer

share|cite|improve this answer

answered Jul 19 at 2:15

Graham Kemp

80.1k43275

answered Jul 19 at 2:15

Graham Kemp

80.1k43275

answered Jul 19 at 2:15

Graham Kemp

80.1k43275

80.1k43275

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