$X$ and $Y$ i.i.d., $X+Y$ and $X-Y$ independent, $mathbb{E}(X)=0 $and $mathbb{E}(X^2)=1$. Show $X sim N(0,1)$












1














$X$ and $Y$ are independent and identically distribued (i.i.d.), $X+Y$ and $X-Y$ are independent, $mathbb{E}(X)=0$ and $mathbb{E}(X^2)=1$. Show that $Xsim N(0,1)$.



We should use characteristic functions to prove this. Any ideas?






probability-theory normal-distribution characteristic-functions







share|cite|improve this question
























  • Sure. And yours?
    – Did
    Nov 7 '13 at 20:41










  • A good reference to someone with a lot of good ideas!? Bogachev! ..in his book Gaussian Measures; particularly section 1.9
    – Tom
    Nov 7 '13 at 20:48










  • What if $X$ and $Y$ are discrete?
    – Michael Hoppe
    Nov 7 '13 at 21:56


















1














$X$ and $Y$ are independent and identically distribued (i.i.d.), $X+Y$ and $X-Y$ are independent, $mathbb{E}(X)=0$ and $mathbb{E}(X^2)=1$. Show that $Xsim N(0,1)$.



We should use characteristic functions to prove this. Any ideas?






probability-theory normal-distribution characteristic-functions







share|cite|improve this question
























  • Sure. And yours?
    – Did
    Nov 7 '13 at 20:41










  • A good reference to someone with a lot of good ideas!? Bogachev! ..in his book Gaussian Measures; particularly section 1.9
    – Tom
    Nov 7 '13 at 20:48










  • What if $X$ and $Y$ are discrete?
    – Michael Hoppe
    Nov 7 '13 at 21:56
















1












1








1


4





$X$ and $Y$ are independent and identically distribued (i.i.d.), $X+Y$ and $X-Y$ are independent, $mathbb{E}(X)=0$ and $mathbb{E}(X^2)=1$. Show that $Xsim N(0,1)$.



We should use characteristic functions to prove this. Any ideas?






probability-theory normal-distribution characteristic-functions







share|cite|improve this question















$X$ and $Y$ are independent and identically distribued (i.i.d.), $X+Y$ and $X-Y$ are independent, $mathbb{E}(X)=0$ and $mathbb{E}(X^2)=1$. Show that $Xsim N(0,1)$.



We should use characteristic functions to prove this. Any ideas?






probability-theory normal-distribution characteristic-functions




probability-theory normal-distribution characteristic-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 16 '14 at 10:21









saz

78.1k758122




78.1k758122










asked Nov 7 '13 at 20:02









user106240

173




173












  • Sure. And yours?
    – Did
    Nov 7 '13 at 20:41










  • A good reference to someone with a lot of good ideas!? Bogachev! ..in his book Gaussian Measures; particularly section 1.9
    – Tom
    Nov 7 '13 at 20:48










  • What if $X$ and $Y$ are discrete?
    – Michael Hoppe
    Nov 7 '13 at 21:56




















  • Sure. And yours?
    – Did
    Nov 7 '13 at 20:41










  • A good reference to someone with a lot of good ideas!? Bogachev! ..in his book Gaussian Measures; particularly section 1.9
    – Tom
    Nov 7 '13 at 20:48










  • What if $X$ and $Y$ are discrete?
    – Michael Hoppe
    Nov 7 '13 at 21:56


















Sure. And yours?
– Did
Nov 7 '13 at 20:41




Sure. And yours?
– Did
Nov 7 '13 at 20:41












A good reference to someone with a lot of good ideas!? Bogachev! ..in his book Gaussian Measures; particularly section 1.9
– Tom
Nov 7 '13 at 20:48




A good reference to someone with a lot of good ideas!? Bogachev! ..in his book Gaussian Measures; particularly section 1.9
– Tom
Nov 7 '13 at 20:48












What if $X$ and $Y$ are discrete?
– Michael Hoppe
Nov 7 '13 at 21:56






What if $X$ and $Y$ are discrete?
– Michael Hoppe
Nov 7 '13 at 21:56












1 Answer
1






active

oldest

votes


















11














Denote by $Phi(t) = mathbb{E}e^{imath , t cdot X}$ the characteristic function of $X$. We have



$$X = frac{1}{2} big((X+Y)+(X-Y) big).$$



Thus,



$$begin{align*} Phi(t) &= mathbb{E}e^{imath , frac{t}{2} (X+Y)} cdot mathbb{E}e^{imath , frac{t}{2} (X-Y)}= left( mathbb{E}e^{imath , frac{t}{2} X} right)^2 cdot mathbb{E}e^{imath , frac{t}{2} Y} cdot mathbb{E}e^{-imath , frac{t}{2} Y} end{align*}$$



where we used the independence of $X-Y$ and $X+Y$ as well as the independence of $X$ and $Y$. By assumption, $X sim Y$; therefore



$$Phi(t) = Phi left( frac{t}{2} right)^3 cdot Phi left( - frac{t}{2} right). tag{1} $$



This shows that it suffices to determine $Phi$ on $B(0,varepsilon)$ for some $varepsilon>0$. Since $Phi(0)=1$, we can choose $varepsilon>0$ such that $Phi(B(0,varepsilon)) cap {x+imath , y; x leq 0, y in mathbb{R}} = emptyset$. For $t in B(0,varepsilon)$ we define



$$psi(t) := log Phi(t)$$



Then $(1)$ reads



$$psi(t) = 3 psi left( frac{t}{2} right) + psi left( - frac{t}{2} right). tag{2} $$



Applying this to $-t$ yields



$$psi(-t) = 3 psi left( - frac{t}{2} right) + psi left( frac{t}{2} right).$$



Subtracting the last two equalities we obtain



$$delta(t) := psi(t)-psi(-t) = 2 psi left( frac{t}{2} right) - 2 psi left( - frac{t}{2} right) = 2 delta left( frac{t}{2} right).$$



Consequently,



$$frac{delta(t)}{t} = frac{delta left( frac{t}{2^n} right)}{frac{t}{2^n}}. tag{3}$$



Note that $Phi$ is twice differentiable and $Phi'(0)=0$, $Phi''(0)=-1$ since $mathbb{E}X=0$, $mathbb{E}(X^2)=1$. Therefore, $delta$ and $psi$ are also twice differentiable and we can calculate the deriatives at $t=0$ explicitely. From $(3)$, we find



$$frac{delta(t)}{t} to delta'(0) = 0qquad text{as} ,, n to infty,$$



i.e. $delta(t)=0$. By the definition of $delta$ and $(2)$,



$$psi(t) = 4 psi left( frac{t}{2} right).$$



By Hôpital's theorem,



$$frac{psi(t)}{t^2} = frac{psi left( frac{t}{2^n} right)}{left( frac{t}{2^n} right)^2} to frac{1}{2} psi''(0) = - frac{1}{2} qquad text{as} ,, n to infty.$$



Hence, $$psi(t) = - frac{t^2}{2}.$$



Reference:




  • Rényi, A.: Probability Theory. (Chapter VI.5 Theorem 1)






share|cite|improve this answer























  • "From Φ(0)=1 we conclude by the intermediate value theorem that Φ(t)>0 for any t∈R"... Except that Φ is complex-valued, not real-valued a priori.
    – Did
    Nov 9 '13 at 1:24










  • @Did You are right. Should be fine now...
    – saz
    Nov 9 '13 at 7:41










  • I fail to see the construction of $(t_n)$. That $Phi(t)^3Phi(-t)$ is in $mathbb R_-$ does not imply a priori that $Phi(t)$ or $Phi(-t)$ is in $mathbb R_-$. // Note that the identity you are trying to use is $Phi(2t)=Phi(t)^2|Phi(t)|^2$.
    – Did
    Nov 9 '13 at 7:50










  • @Did Yes, of course, sorry.
    – saz
    Nov 9 '13 at 9:33










  • Rrrright... +1.
    – Did
    Nov 9 '13 at 9:45











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









11














Denote by $Phi(t) = mathbb{E}e^{imath , t cdot X}$ the characteristic function of $X$. We have



$$X = frac{1}{2} big((X+Y)+(X-Y) big).$$



Thus,



$$begin{align*} Phi(t) &= mathbb{E}e^{imath , frac{t}{2} (X+Y)} cdot mathbb{E}e^{imath , frac{t}{2} (X-Y)}= left( mathbb{E}e^{imath , frac{t}{2} X} right)^2 cdot mathbb{E}e^{imath , frac{t}{2} Y} cdot mathbb{E}e^{-imath , frac{t}{2} Y} end{align*}$$



where we used the independence of $X-Y$ and $X+Y$ as well as the independence of $X$ and $Y$. By assumption, $X sim Y$; therefore



$$Phi(t) = Phi left( frac{t}{2} right)^3 cdot Phi left( - frac{t}{2} right). tag{1} $$



This shows that it suffices to determine $Phi$ on $B(0,varepsilon)$ for some $varepsilon>0$. Since $Phi(0)=1$, we can choose $varepsilon>0$ such that $Phi(B(0,varepsilon)) cap {x+imath , y; x leq 0, y in mathbb{R}} = emptyset$. For $t in B(0,varepsilon)$ we define



$$psi(t) := log Phi(t)$$



Then $(1)$ reads



$$psi(t) = 3 psi left( frac{t}{2} right) + psi left( - frac{t}{2} right). tag{2} $$



Applying this to $-t$ yields



$$psi(-t) = 3 psi left( - frac{t}{2} right) + psi left( frac{t}{2} right).$$



Subtracting the last two equalities we obtain



$$delta(t) := psi(t)-psi(-t) = 2 psi left( frac{t}{2} right) - 2 psi left( - frac{t}{2} right) = 2 delta left( frac{t}{2} right).$$



Consequently,



$$frac{delta(t)}{t} = frac{delta left( frac{t}{2^n} right)}{frac{t}{2^n}}. tag{3}$$



Note that $Phi$ is twice differentiable and $Phi'(0)=0$, $Phi''(0)=-1$ since $mathbb{E}X=0$, $mathbb{E}(X^2)=1$. Therefore, $delta$ and $psi$ are also twice differentiable and we can calculate the deriatives at $t=0$ explicitely. From $(3)$, we find



$$frac{delta(t)}{t} to delta'(0) = 0qquad text{as} ,, n to infty,$$



i.e. $delta(t)=0$. By the definition of $delta$ and $(2)$,



$$psi(t) = 4 psi left( frac{t}{2} right).$$



By Hôpital's theorem,



$$frac{psi(t)}{t^2} = frac{psi left( frac{t}{2^n} right)}{left( frac{t}{2^n} right)^2} to frac{1}{2} psi''(0) = - frac{1}{2} qquad text{as} ,, n to infty.$$



Hence, $$psi(t) = - frac{t^2}{2}.$$



Reference:




  • Rényi, A.: Probability Theory. (Chapter VI.5 Theorem 1)






share|cite|improve this answer























  • "From Φ(0)=1 we conclude by the intermediate value theorem that Φ(t)>0 for any t∈R"... Except that Φ is complex-valued, not real-valued a priori.
    – Did
    Nov 9 '13 at 1:24










  • @Did You are right. Should be fine now...
    – saz
    Nov 9 '13 at 7:41










  • I fail to see the construction of $(t_n)$. That $Phi(t)^3Phi(-t)$ is in $mathbb R_-$ does not imply a priori that $Phi(t)$ or $Phi(-t)$ is in $mathbb R_-$. // Note that the identity you are trying to use is $Phi(2t)=Phi(t)^2|Phi(t)|^2$.
    – Did
    Nov 9 '13 at 7:50










  • @Did Yes, of course, sorry.
    – saz
    Nov 9 '13 at 9:33










  • Rrrright... +1.
    – Did
    Nov 9 '13 at 9:45
















11














Denote by $Phi(t) = mathbb{E}e^{imath , t cdot X}$ the characteristic function of $X$. We have



$$X = frac{1}{2} big((X+Y)+(X-Y) big).$$



Thus,



$$begin{align*} Phi(t) &= mathbb{E}e^{imath , frac{t}{2} (X+Y)} cdot mathbb{E}e^{imath , frac{t}{2} (X-Y)}= left( mathbb{E}e^{imath , frac{t}{2} X} right)^2 cdot mathbb{E}e^{imath , frac{t}{2} Y} cdot mathbb{E}e^{-imath , frac{t}{2} Y} end{align*}$$



where we used the independence of $X-Y$ and $X+Y$ as well as the independence of $X$ and $Y$. By assumption, $X sim Y$; therefore



$$Phi(t) = Phi left( frac{t}{2} right)^3 cdot Phi left( - frac{t}{2} right). tag{1} $$



This shows that it suffices to determine $Phi$ on $B(0,varepsilon)$ for some $varepsilon>0$. Since $Phi(0)=1$, we can choose $varepsilon>0$ such that $Phi(B(0,varepsilon)) cap {x+imath , y; x leq 0, y in mathbb{R}} = emptyset$. For $t in B(0,varepsilon)$ we define



$$psi(t) := log Phi(t)$$



Then $(1)$ reads



$$psi(t) = 3 psi left( frac{t}{2} right) + psi left( - frac{t}{2} right). tag{2} $$



Applying this to $-t$ yields



$$psi(-t) = 3 psi left( - frac{t}{2} right) + psi left( frac{t}{2} right).$$



Subtracting the last two equalities we obtain



$$delta(t) := psi(t)-psi(-t) = 2 psi left( frac{t}{2} right) - 2 psi left( - frac{t}{2} right) = 2 delta left( frac{t}{2} right).$$



Consequently,



$$frac{delta(t)}{t} = frac{delta left( frac{t}{2^n} right)}{frac{t}{2^n}}. tag{3}$$



Note that $Phi$ is twice differentiable and $Phi'(0)=0$, $Phi''(0)=-1$ since $mathbb{E}X=0$, $mathbb{E}(X^2)=1$. Therefore, $delta$ and $psi$ are also twice differentiable and we can calculate the deriatives at $t=0$ explicitely. From $(3)$, we find



$$frac{delta(t)}{t} to delta'(0) = 0qquad text{as} ,, n to infty,$$



i.e. $delta(t)=0$. By the definition of $delta$ and $(2)$,



$$psi(t) = 4 psi left( frac{t}{2} right).$$



By Hôpital's theorem,



$$frac{psi(t)}{t^2} = frac{psi left( frac{t}{2^n} right)}{left( frac{t}{2^n} right)^2} to frac{1}{2} psi''(0) = - frac{1}{2} qquad text{as} ,, n to infty.$$



Hence, $$psi(t) = - frac{t^2}{2}.$$



Reference:




  • Rényi, A.: Probability Theory. (Chapter VI.5 Theorem 1)






share|cite|improve this answer























  • "From Φ(0)=1 we conclude by the intermediate value theorem that Φ(t)>0 for any t∈R"... Except that Φ is complex-valued, not real-valued a priori.
    – Did
    Nov 9 '13 at 1:24










  • @Did You are right. Should be fine now...
    – saz
    Nov 9 '13 at 7:41










  • I fail to see the construction of $(t_n)$. That $Phi(t)^3Phi(-t)$ is in $mathbb R_-$ does not imply a priori that $Phi(t)$ or $Phi(-t)$ is in $mathbb R_-$. // Note that the identity you are trying to use is $Phi(2t)=Phi(t)^2|Phi(t)|^2$.
    – Did
    Nov 9 '13 at 7:50










  • @Did Yes, of course, sorry.
    – saz
    Nov 9 '13 at 9:33










  • Rrrright... +1.
    – Did
    Nov 9 '13 at 9:45














11












11








11






Denote by $Phi(t) = mathbb{E}e^{imath , t cdot X}$ the characteristic function of $X$. We have



$$X = frac{1}{2} big((X+Y)+(X-Y) big).$$



Thus,



$$begin{align*} Phi(t) &= mathbb{E}e^{imath , frac{t}{2} (X+Y)} cdot mathbb{E}e^{imath , frac{t}{2} (X-Y)}= left( mathbb{E}e^{imath , frac{t}{2} X} right)^2 cdot mathbb{E}e^{imath , frac{t}{2} Y} cdot mathbb{E}e^{-imath , frac{t}{2} Y} end{align*}$$



where we used the independence of $X-Y$ and $X+Y$ as well as the independence of $X$ and $Y$. By assumption, $X sim Y$; therefore



$$Phi(t) = Phi left( frac{t}{2} right)^3 cdot Phi left( - frac{t}{2} right). tag{1} $$



This shows that it suffices to determine $Phi$ on $B(0,varepsilon)$ for some $varepsilon>0$. Since $Phi(0)=1$, we can choose $varepsilon>0$ such that $Phi(B(0,varepsilon)) cap {x+imath , y; x leq 0, y in mathbb{R}} = emptyset$. For $t in B(0,varepsilon)$ we define



$$psi(t) := log Phi(t)$$



Then $(1)$ reads



$$psi(t) = 3 psi left( frac{t}{2} right) + psi left( - frac{t}{2} right). tag{2} $$



Applying this to $-t$ yields



$$psi(-t) = 3 psi left( - frac{t}{2} right) + psi left( frac{t}{2} right).$$



Subtracting the last two equalities we obtain



$$delta(t) := psi(t)-psi(-t) = 2 psi left( frac{t}{2} right) - 2 psi left( - frac{t}{2} right) = 2 delta left( frac{t}{2} right).$$



Consequently,



$$frac{delta(t)}{t} = frac{delta left( frac{t}{2^n} right)}{frac{t}{2^n}}. tag{3}$$



Note that $Phi$ is twice differentiable and $Phi'(0)=0$, $Phi''(0)=-1$ since $mathbb{E}X=0$, $mathbb{E}(X^2)=1$. Therefore, $delta$ and $psi$ are also twice differentiable and we can calculate the deriatives at $t=0$ explicitely. From $(3)$, we find



$$frac{delta(t)}{t} to delta'(0) = 0qquad text{as} ,, n to infty,$$



i.e. $delta(t)=0$. By the definition of $delta$ and $(2)$,



$$psi(t) = 4 psi left( frac{t}{2} right).$$



By Hôpital's theorem,



$$frac{psi(t)}{t^2} = frac{psi left( frac{t}{2^n} right)}{left( frac{t}{2^n} right)^2} to frac{1}{2} psi''(0) = - frac{1}{2} qquad text{as} ,, n to infty.$$



Hence, $$psi(t) = - frac{t^2}{2}.$$



Reference:




  • Rényi, A.: Probability Theory. (Chapter VI.5 Theorem 1)






share|cite|improve this answer














Denote by $Phi(t) = mathbb{E}e^{imath , t cdot X}$ the characteristic function of $X$. We have



$$X = frac{1}{2} big((X+Y)+(X-Y) big).$$



Thus,



$$begin{align*} Phi(t) &= mathbb{E}e^{imath , frac{t}{2} (X+Y)} cdot mathbb{E}e^{imath , frac{t}{2} (X-Y)}= left( mathbb{E}e^{imath , frac{t}{2} X} right)^2 cdot mathbb{E}e^{imath , frac{t}{2} Y} cdot mathbb{E}e^{-imath , frac{t}{2} Y} end{align*}$$



where we used the independence of $X-Y$ and $X+Y$ as well as the independence of $X$ and $Y$. By assumption, $X sim Y$; therefore



$$Phi(t) = Phi left( frac{t}{2} right)^3 cdot Phi left( - frac{t}{2} right). tag{1} $$



This shows that it suffices to determine $Phi$ on $B(0,varepsilon)$ for some $varepsilon>0$. Since $Phi(0)=1$, we can choose $varepsilon>0$ such that $Phi(B(0,varepsilon)) cap {x+imath , y; x leq 0, y in mathbb{R}} = emptyset$. For $t in B(0,varepsilon)$ we define



$$psi(t) := log Phi(t)$$



Then $(1)$ reads



$$psi(t) = 3 psi left( frac{t}{2} right) + psi left( - frac{t}{2} right). tag{2} $$



Applying this to $-t$ yields



$$psi(-t) = 3 psi left( - frac{t}{2} right) + psi left( frac{t}{2} right).$$



Subtracting the last two equalities we obtain



$$delta(t) := psi(t)-psi(-t) = 2 psi left( frac{t}{2} right) - 2 psi left( - frac{t}{2} right) = 2 delta left( frac{t}{2} right).$$



Consequently,



$$frac{delta(t)}{t} = frac{delta left( frac{t}{2^n} right)}{frac{t}{2^n}}. tag{3}$$



Note that $Phi$ is twice differentiable and $Phi'(0)=0$, $Phi''(0)=-1$ since $mathbb{E}X=0$, $mathbb{E}(X^2)=1$. Therefore, $delta$ and $psi$ are also twice differentiable and we can calculate the deriatives at $t=0$ explicitely. From $(3)$, we find



$$frac{delta(t)}{t} to delta'(0) = 0qquad text{as} ,, n to infty,$$



i.e. $delta(t)=0$. By the definition of $delta$ and $(2)$,



$$psi(t) = 4 psi left( frac{t}{2} right).$$



By Hôpital's theorem,



$$frac{psi(t)}{t^2} = frac{psi left( frac{t}{2^n} right)}{left( frac{t}{2^n} right)^2} to frac{1}{2} psi''(0) = - frac{1}{2} qquad text{as} ,, n to infty.$$



Hence, $$psi(t) = - frac{t^2}{2}.$$



Reference:




  • Rényi, A.: Probability Theory. (Chapter VI.5 Theorem 1)







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Apr 19 '15 at 6:52

























answered Nov 8 '13 at 16:53









saz

78.1k758122




78.1k758122












  • "From Φ(0)=1 we conclude by the intermediate value theorem that Φ(t)>0 for any t∈R"... Except that Φ is complex-valued, not real-valued a priori.
    – Did
    Nov 9 '13 at 1:24










  • @Did You are right. Should be fine now...
    – saz
    Nov 9 '13 at 7:41










  • I fail to see the construction of $(t_n)$. That $Phi(t)^3Phi(-t)$ is in $mathbb R_-$ does not imply a priori that $Phi(t)$ or $Phi(-t)$ is in $mathbb R_-$. // Note that the identity you are trying to use is $Phi(2t)=Phi(t)^2|Phi(t)|^2$.
    – Did
    Nov 9 '13 at 7:50










  • @Did Yes, of course, sorry.
    – saz
    Nov 9 '13 at 9:33










  • Rrrright... +1.
    – Did
    Nov 9 '13 at 9:45


















  • "From Φ(0)=1 we conclude by the intermediate value theorem that Φ(t)>0 for any t∈R"... Except that Φ is complex-valued, not real-valued a priori.
    – Did
    Nov 9 '13 at 1:24










  • @Did You are right. Should be fine now...
    – saz
    Nov 9 '13 at 7:41










  • I fail to see the construction of $(t_n)$. That $Phi(t)^3Phi(-t)$ is in $mathbb R_-$ does not imply a priori that $Phi(t)$ or $Phi(-t)$ is in $mathbb R_-$. // Note that the identity you are trying to use is $Phi(2t)=Phi(t)^2|Phi(t)|^2$.
    – Did
    Nov 9 '13 at 7:50










  • @Did Yes, of course, sorry.
    – saz
    Nov 9 '13 at 9:33










  • Rrrright... +1.
    – Did
    Nov 9 '13 at 9:45
















"From Φ(0)=1 we conclude by the intermediate value theorem that Φ(t)>0 for any t∈R"... Except that Φ is complex-valued, not real-valued a priori.
– Did
Nov 9 '13 at 1:24




"From Φ(0)=1 we conclude by the intermediate value theorem that Φ(t)>0 for any t∈R"... Except that Φ is complex-valued, not real-valued a priori.
– Did
Nov 9 '13 at 1:24












@Did You are right. Should be fine now...
– saz
Nov 9 '13 at 7:41




@Did You are right. Should be fine now...
– saz
Nov 9 '13 at 7:41












I fail to see the construction of $(t_n)$. That $Phi(t)^3Phi(-t)$ is in $mathbb R_-$ does not imply a priori that $Phi(t)$ or $Phi(-t)$ is in $mathbb R_-$. // Note that the identity you are trying to use is $Phi(2t)=Phi(t)^2|Phi(t)|^2$.
– Did
Nov 9 '13 at 7:50




I fail to see the construction of $(t_n)$. That $Phi(t)^3Phi(-t)$ is in $mathbb R_-$ does not imply a priori that $Phi(t)$ or $Phi(-t)$ is in $mathbb R_-$. // Note that the identity you are trying to use is $Phi(2t)=Phi(t)^2|Phi(t)|^2$.
– Did
Nov 9 '13 at 7:50












@Did Yes, of course, sorry.
– saz
Nov 9 '13 at 9:33




@Did Yes, of course, sorry.
– saz
Nov 9 '13 at 9:33












Rrrright... +1.
– Did
Nov 9 '13 at 9:45




Rrrright... +1.
– Did
Nov 9 '13 at 9:45


















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