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Let $X$ be a uniformly chosen number in the interval $(0, 1)$. Choose a point P uniformly from the triangle with vertices at $(X,0), (0,X)$, and $(0,0)$. Let $Y$ be the $y$ coordinate of the point P . Compute $E(Y )$.

Here is my work:

enter image description here

Here is what I know:

$E(Y) = \int_{-\infty}^{\infty}E(Y|X=x)\ * \ f_x(x)\ dx$

and

$E(Y|X=x) = \int_{-\infty}^{\infty}y * f_{Y|X}(y|x)\ dy$

So since, the blue line can be expressed as $y = X - x$, then the boundaries are $0<y<X-x$ and $0<x<1$

Then I said:

\begin{equation} E(Y)=\int_{0}^{1}\int_{0}^{X-x}y*f_{Y|X}(y|x)*f_{X}(x)\ dy\ dx \end{equation}

However, I'm stuck since, I don't know how to find $f_{Y|X}(y|x)$

Am I doing this right? If so, how do I find $f_{Y|X}(y|x)$ so I can complete the problem?

user3904534
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    The boundary $0 < y < X-x$ does not make sense, neither does big $X$ appearing in your expression for $E[Y]$. In fact, given $X=x$, we know that $y \in [0,x]$, so these would be the correct boundaries. – Michael Dec 02 '15 at 04:30
  • Perhaps what is confusing you is your label for the horizontal axis, you cannot label this $x$ since you are already using that. So you could label it $z$. Then, given $X=x$ and we have a point $(Z,Y)$ on the triangle defined by $X=x$, we get $f_Y(y) = \int_{z=-\infty}^{\infty} f_{Z,Y}(z,y)dz$ (where these are all implicitly conditioned on $X=x$). – Michael Dec 02 '15 at 04:32

2 Answers2

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When $X=x$, the point $P$ is uniformly distributed in the triangle $\triangle(0,0)(0,x)(x,0)$ so the density of $Y$ at $y$ is measured by the length of the horizontal line-segment at height $y$ passing through the triangle divided by the area of the triangle.

$$f_{Y\mid X}(y\mid x) = \frac{2(x-y)}{x^2}\quad \mathbf 1_{y\in[0;x]}$$

Then: $\mathsf E(Y)$ $ \displaystyle = \int_0^1\int_0^x y f_{Y\mid X}(y\mid x)f_X(x)\operatorname d y\operatorname d x \\\displaystyle = \int_0^1\int_0^x \frac{2y(x-y)}{x^2}\operatorname d y\operatorname d x$

Graham Kemp
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We find $E(Y)$, given that $X=x$. One slightly klunky way of doing this is to find the conditional distribution of $Y$ given that $X=x$.

Let $y\le x$. Then the probability that $Y\gt y$ is the area of the part of the triangle above the horizontal line at height $y$. This triangle has legs $x-y$, so area $(1/2)(x-y)^2$. So the conditional probability that $Y\gt y$ is $\frac{(x-y)^2/2}{x^2/2}$. It follows that the conditional density of $Y$ is $\frac{2(x-y)}{x^2}$. The conditional expectation of $Y$ is therefore $$\int_0^x \frac{2y(x-y)}{x^2}\,dx.$$ This is $\frac{2}{3}x$. For the expectation of $Y$, calculate $\int_0^1 \frac{2}{3}x\,dx$.

André Nicolas
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