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Say we have two $i.i.d.$ random variables $X$ and $Y$ with support in $[0,1]$. We want to find

$E[X| X > 0.7, Y > 0.7, \frac{X+Y}{2} > 0.9]$. This is given by:

$\int x f(x|x > 0.7, y > 0.7, \frac{x+y}{2} > 0.9) dX $.

What will be the expression for $f(x|x > 0.7, y > 0.7, \frac{x+y}{2} > 0.9)$?

Also is there any book or material where I can refer these type of examples ?

Pradipta
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1 Answers1

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In the square $[0,1]^2$, both conditions $x\gt0.7$ and $y\gt0.7$ are implied by the condition $\frac12(x+y)\gt0.9$ hence $T=\{(x,y)\in[0,1]^2\mid x\gt0.7,y\gt0.7,\frac12(x+y)\gt0.9\}$ is actually the triangle with vertices $(0.8,1)$, $(1,1)$ and $(1,0.8)$. The conditional density of $(X,Y)$ conditionally on $(X,Y)\in T$ is constant hence the conditional density of $X$ conditionally on $(X,Y)\in T$ is proportional to $(x-0.8)\cdot\mathbf 1_{0.8\leqslant x\leqslant1}$ and $$ E(X\mid (X,Y)\in T)=\frac{\displaystyle\int_{0.8}^1x(x-0.8)\,\mathrm dx}{\displaystyle\int_{0.8}^1(x-0.8)\,\mathrm dx}=\frac{\displaystyle\int_0^{0.2}t(t+0.8)\,\mathrm dt}{\displaystyle\int_0^{0.2}t\,\mathrm dt}=\text{____}. $$ The conditional distribution $f_{X|A}$ of $X$ conditionally on $A=[(X,Y)\in T]$ is $$ f_{X|A}(x)=50\cdot(x-0.8)\cdot\mathbf 1_{0.8\leqslant x\leqslant1}. $$ Textbook: basically any text dealing with continuous distributions and conditional expectations (plethora of them on the web). Which ones are suitable to you depends on your background.

Edit: If $X$ is an integrable random variable and $(B_n)$ are disjoint events with union $B$, then $$ E(X\mid B)=\frac{E(X;B)}{P(B)}=\frac{\sum\limits_nE(X;B_n)}{\sum\limits_nP(B_n)}=\frac{\sum\limits_nE(X\mid B_n)P(B_n)}{\sum\limits_nP(B_n)}. $$ If there are $N$ events $B_n$ and if $P(B_n)$ does not depend on $n$, it follows that $$ E(X\mid B)=\frac1N\sum\limits_nE(X\mid B_n). $$ Even in this case, $E(X\mid B)\ne\sum\limits_nE(X\mid B_n)$, some kind of weird algebraic coincidence excepted.

Did
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  • Thanks for the quick response. I had two followup questions. – Pradipta Feb 02 '14 at 13:18
  • Thanks for the quick response. I had two followup questions. (a) You have used uniform distribution for your analysis ? (b) $f(x|x>0.7, y>0.7, (x+y)/2 > 0.9) = \frac{\int_{1.8-x}^{1}f(x,y)dy}{\int_{0.8}^{1}\int_{1.8-x}^{1}f(x,y)dy dx}$. This is the general form that you have used ? – Pradipta Feb 02 '14 at 13:26
  • (a) Of course. (b) Provided one gets rid of these irritating notations confusing parameters and integrands, yes. – Did Feb 02 '14 at 14:17
  • I had another followup question. Can you please confirm whether I am right. I want to find $E(X|(X,Y)\in T_{1})$ where $T_{1}={(x,y) \in [0,1]^2|x>0.7, y>0.7, \frac{1}{2}(x+y)< 0.9}$. So $T_{1}$ gives a pentagon with vertices $(0.7,1)$, $(0.8,1)$, $(1,0.8)$, $(1,0.7)$ and $(0.7,0.7)$. This region does not have a constant shape. So we split it into three regions: $T_{11}$ with vertices $(0.7,1)$, $(0.8,1)$, $(0.8,0.7)$ and $(0.7,0.7)$; $T_{12}$ with vertices $(0.8,0.8)$, $(1,0.8)$, $(1,0.7)$ and $(0.8,0.7)$; $T_{13}$ with vertices $(0.8,1), (0.8,0.8)$ and $(1,0.8)$. – Pradipta Feb 04 '14 at 02:31
  • Continuing with my above comment since there was no space left. Now to find $E(X|(X,Y) \in T_{1})$, can we just do the following ? $E(X|(X,Y) \in T_{11})$ + $E(X|(X,Y) \in T_{12})$ + $E(X|(X,Y) \in T_{13})$ i.e. integrate over the three sub regions and add. – Pradipta Feb 04 '14 at 02:33
  • Reading your last comments, I am a bit worried by the depth of your grasp of conditional expectations. Do you even know the definition? Anyway, see Edit. – Did Feb 04 '14 at 07:23
  • I agree I made a blunder with my above comment. My main worry was how to find the conditional distribution over an area which has two rectangles and a triangle ? – Pradipta Feb 04 '14 at 08:56
  • ok. I think I got some idea how to evaluate this. $E(X|(X,Y) \in T_{1})$ = $\int_{0.7}^{0.8} x f_{X}(x|A)dx$ + $\int_{0.8}^{1} x f_{X}(x|B)dx$ + $\int_{0.8}^{1} x f_{X}(x|C)dx$. $A,B,C$ represents the constraints of three sub-regions $T_{11}, T_{12}, T_{13}$. – Pradipta Feb 04 '14 at 09:17
  • Once again, no, $E(X\mid(X,Y)\in T_1)$ is NOT what you write. (Did you read my Edit?) – Did Feb 04 '14 at 09:39
  • ok. You mean, I have to use this: $$ E(X\mid B)=\frac1N\sum\limits_nE(X\mid B_n). $$ – Pradipta Feb 04 '14 at 10:30
  • Certainly not, since in your case $P(T_{1i})$ depends on $i$. – Did Feb 04 '14 at 10:33
  • or the more general form $\frac{\sum\limits_nE(X\mid B_n)P(B_n)}{\sum\limits_nP(B_n)}$. – Pradipta Feb 04 '14 at 10:34
  • As explained in my post. But this decomposition probably only makes things more complicated... – Did Feb 04 '14 at 10:37
  • Here the decomposition is needed because the concerned region $T_{1}$ does not have a constant shape ? – Pradipta Feb 04 '14 at 10:48
  • Is there any other simpler way of computing other than decomposition ? – Pradipta Feb 04 '14 at 13:44
  • Sure--for example by substracting the integral on the upper right triangle from the integral on the square. – Did Feb 04 '14 at 14:00
  • I can simply do this: $E(X)$ = $E(X|(X,Y) \in T)P((X,Y) \in T)$ + $E(X|(X,Y) \in T_{1})P((X,Y) \in T_{1})$. – Pradipta Feb 04 '14 at 14:45