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$X, Y$ iid uniform random variables on $[0,1]$ $Z = \left\{ \begin{aligned} X+Y \quad&\text{ if } X>\frac{1}{2} \\ \frac{1}{2} + Y \quad & \text{ if } X\leq\frac{1}{2} \end{aligned} \right.$ The question is $E\{Z|Z\leq 1\}= ?$

I tried $\displaystyle \int_0^1 E\{Z|Z = z\} P\{Z = z\}dz$ and got $5/8$, but I am not so sure about the result since I haven't touched probability for years.

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    thanx for the heads up, I forgot to put the density function for $Z$2011-07-19

2 Answers 2

5

For $x \in (1/2,1]$, let $ F_{Z|Z \leq 1}(x) = {\rm P}(Z \leq x | Z \leq 1) = \frac{{{\rm P}(Z \le x)}}{{{\rm P}(Z \le 1)}}, $ and let $ f_{Z|Z \leq 1}(x) = \frac{{\rm d}}{{{\rm d}x}}F_{Z|Z \le 1} (x) = \frac{1}{{{\rm P}(Z \le 1)}}\frac{{\rm d}}{{{\rm d}x}}{\rm P}(Z \le x). $ Then, $ {\rm E}[Z|Z \le 1] = \int_{1/2}^1 {xf_{Z|Z \le 1} (x)\,{\rm d}x} . $ So now the problem reduces to calculating ${\rm P}(Z \leq x)$, for $x \in (1/2,1]$. This can be done using the law of total probability, conditioning on $X$, leading to $ {\rm E}[Z|Z \le 1] = 7/9. $

EDIT:

Fix $x \in (1/2,1]$. Then, by the law of total probability, $ {\rm P}(Z \leq x) = \int_0^{1/2} {{\rm P}(Z \le x|X = s)\,{\rm d}s} + \int_{1/2}^1 {{\rm P}(Z \le x|X = s)\,{\rm d}s} . $ It thus follows from the definition of $Z$ (and the independence of $X$ and $Y$) that $ {\rm P}(Z \leq x) = \int_0^{1/2} {{\rm P}( Y \le x - 1/2)\,{\rm d}s} + \int_{1/2}^1 {{\rm P}(Y \le x - s)\,{\rm d}s} . $ Now, $ \int_0^{1/2} {{\rm P}( Y \le x - 1/2)\,{\rm d}s} = \frac{1}{2}{\rm P}(Y \le x - 1/2) = \frac{{x - 1/2}}{2} = \frac{x}{2} - \frac{1}{4} $ and $ \int_{1/2}^1 {{\rm P}(Y \le x - s)\,{\rm d}s} = \int_{1/2}^x {{\rm P}(Y \le x - s)\,{\rm d}s} = \int_{1/2}^x {(x - s)\,{\rm d}s} = \frac{{x^2 }}{2} - \frac{x}{2} + \frac{1}{8}. $ Hence $ {\rm P}(Z \leq x) = \bigg(\frac{x}{2} - \frac{1}{4}\bigg) + \bigg(\frac{{x^2 }}{2} - \frac{x}{2} + \frac{1}{8}\bigg) = \frac{{x^2 }}{2} - \frac{1}{8}. $ In particular, $ {\rm P}(Z \leq 1) = \frac{3}{8}. $ Thus, $ F_{Z|Z \leq 1}(x) = \frac{8}{3}\bigg(\frac{{x^2 }}{2} - \frac{1}{8}\bigg) = \frac{{4x^2 - 1}}{3}, $ and in turn $ f_{Z|Z \leq 1}(x) = \frac{{8x}}{3}. $ Finally, $ {\rm E}[Z|Z \le 1] = \int_{1/2}^1 {xf_{Z|Z \le 1} (x)\,{\rm d}x} = \frac{8}{3}\int_{1/2}^1 {x^2 \,{\rm d}x} = \frac{8}{3}\bigg(\frac{{1 - 1/8}}{3}\bigg) = \frac{7}{9}. $

5

Your probability space is the unit square in the $(x,y)$-plane with $dP={\rm d}(x,y)$. The payout $Z$ is ${1\over 2}+y$ in the left half $L$ of the square and $x+y$ in the right half $R$. The region where $Z\leq 1$ consists of the lower half of $L$ and a triangle in the lower left of $R$; it has total area $P(Z\leq 1)={3\over8}$.

It follows that the expectation $E:=E[Z\ |\ Z\leq 1]$ is given by $E=\left(\int_0^{1/2}\int_0^{1/2}\bigl({1\over2}+y\bigr)dy dx + \int_{1/2}^1\int_0^{1-x}(x+y)dy dx\right)\Bigg/{3\over8} ={{3\over16}+{5\over48}\over{3\over8}}={7\over9}\ .$