# Joint probability density function probability

$X$ and $Y$ are jointly continuous random variables.

$$f(x,y)=\begin{cases}kx & x\in[0,1], y\in [0,1]\\0 & :\text{otherwise}\end{cases}$$

a) What value of $k$ makes this a density function?

I have calculated a value of $2$ using the double integral, is this correct?

b) Calculate $\mathsf P(2X > Y )$.

I am unsure what to do here.

• Your $k$ is correct. For the other problem, draw the square, draw the line $y=2x$. You want the probability of being in the square but below the line. That's an integral over a slightly funny region, you may have to break it up. Or find first the probability of the complement. – André Nicolas Oct 29 '14 at 19:10
• So I got 0.9167, is that correct? – StatsHelp Oct 29 '14 at 19:52
• Yes, I got $\frac{11}{12}$, which to $4$ decimal places is your number. – André Nicolas Oct 29 '14 at 20:29

a) $k=2$ is correct.
b) Prescribe function $g$ on $[0,1]^2$ by $(x,y)\mapsto 1$ if $2x>y$ and $(x,y)\mapsto 0$ otherwise.
Now find $\mathbb P(2X>Y)=\mathbb Eg(X,Y)$ by evaluating the integral $$\int_0^1\int_0^1 f(x,y)g(x,y)dxdy=2\int_0^1\int_0^1 xg(x,y)dxdy$$
Here: $$\int_0^1\int_0^1 xg(x,y)dxdy=\int_0^1\int_0^1 xg(x,y)dydx$$ so make your choice.
• @StatsHelp: \begin{align} \mathsf P(2X > Y) & = \int_0^{1/2} \int_{0}^{2x} 2x \operatorname d y\operatorname d x + \int_{1/2}^1\int_0^1 2x\operatorname d y \operatorname d x \\[1ex] & = \frac {11}{12} & \checkmark \end{align} – Graham Kemp Oct 29 '14 at 23:29