Conditional distribution of uniform (0,1).

Question

Two independent random variables are distributed as $X \sim R(0,1)$ and $ Y \sim R(0,1)$.

My problem: $P\big(X+Y>1|Y>\frac{1}{2}\big)$=?

I know that when a random variable is uniformly distributed that $F(x)=P(X\leq x)=\frac{x-a}{b-a} , x\in]a,b[$ which gives me $F(x)=\frac{x-0}{1-0}$, since its distributed on the interval $(0,1)$. Also i know that the density, f(x) is equal to $f(x)=\frac{1}{b-a}$, which gives me $f(y)=f(x)=\frac{1}{1-0}=1$

Since they are independent: $f(x)\cdot f(y)=f(x,y) \Rightarrow f(x,y)=1\cdot 1=1$

I get: $P(X+Y>1)=\int_0^1\int_{1-y}^11 \ dxdy=\frac{1}{2}$

and

$P(Y>\frac{1}{2})=1-\frac{\frac{1}{2}-0}{1-0}=\frac{1}{2}$

I use the formula: $P(A|B)=\frac{P(A \cap B)}{P(B)} \Rightarrow \frac{\frac{1}{2} \cdot \frac{1}{2}}{\frac{1}{2}}=\frac{1}{2}$ which is wrong.

Can anyone help me solving this?

Answer 1

The density of the sum of independent random variables is a convolution of their respective densities, not the product.

Here, however, you can even avoid the convolution by using the geometric approach. Draw a square $[0,1]\times [0,1]$, and find which part of this square corresponds to the set $\{x+y>1,\,y>1/2\}$. Since the densities are uniform, the probability $P(\{X+Y>1\} \cap \{Y>1/2\})$ is equal to the measure of the set you've just found.

Answer 2

If you say $$\frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{2} \cdot \frac{1}{2}}{\frac{1}{2}}$$ you imply $$P(A \cap B) = \frac{1}{2} \cdot \frac{1}{2} = P(A)P(B)$$ so you actually assume that $$A = \{ X + Y > 1\}$$ and $$ B = \{ Y > \frac{1}{2}\}$$ are independent what they aren't.

You just missed to calculate $$P(A \cap B) = P\left( X + Y > 1, Y > \frac{1}{2}\right)$$ directly what can easily be done by the joint distribution $$\begin{align*} P\left( X + Y > 1, Y > \frac{1}{2}\right) &= \int_0^1 \int_0^1 1_{\{X + Y > 1, Y > \frac{1}{2}\}} \;f(x,y) dx dy \\ &= \int_\frac{1}{2}^1 \int_{1-y}^1 1 \ dx dy \\&= \frac{1}{2}\cdot \frac{3}{4}\end{align*}$$

So you get: $$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{2}\cdot \frac{3}{4}}{\frac{1}{2}} = \frac{3}{4}$$