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If the joint probability density of $X$ and $Y$ is given by $$f(x,y)=2 $$ for $x>0,y>0,x+y<1$

Find $P(X>2Y)$

Can anyone give me some hints for this questions? Thanks in advance.

Mathxx
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    You can consider this a hint, as it is not exactly the same question, but is very very similar: https://math.stackexchange.com/questions/2538705/joint-probability-density-function-and-limits-of-integration/2538757#2538757 Have a read of the answer to see the method used. – John Doe Nov 29 '17 at 16:32

3 Answers3

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Maybe it is better to consider the figure

enter image description here

The area of the red triangles is $\frac16$. The value of the joint density is constant $2$. So, the probability we are after is $$P=\frac13.$$

zoli
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$$P(X<2Y) = \int_{0}^{\frac{2}{3}}\int_{0}^{\frac{x}{2}}2dydx + \int_{\frac{2}{3}}^{1}\int_{0}^{1-x}2dydx =\frac{1}{3}$$

Goodness, @zoli has added the diagram. From this it is clear that the breakpoint in the integral is $2/3$ for x and y from $0$ to $x/2$ till $2/3$ and from $0$ to $1-x$ from $2/3$ to $1$.

Satish Ramanathan
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  • Can you explain your solution for me? Thanks a lot. – Mathxx Nov 29 '17 at 16:55
  • @Mathxx It is an area integral of $f(x,y)$, hence why there is a $2$ in the integral. The region over which you integrate is the region where $X$ is less than $2Y$, ie. under the line $y=\frac x2$. This explains the limits. Edit: The other answer has a sketch of this region. – John Doe Nov 29 '17 at 16:57
  • Thanks @John Doe, I have given a simple answer. I don't have a graphing software sorry. – Satish Ramanathan Nov 29 '17 at 17:01
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This may be an even simpler interpretation for any future readers.

When working with inequalities like $P(X > aY)$ for Probability Density Functions (PDFs)

After having drawn out a few examples on desmos for $P(X > 2Y)$,$P(X > 3Y)$, $P(X > 4Y)$.

I have found a pattern at the the point of intersection, for example when $P(X>2Y)$ the point of intersection is $(\frac{2}{3}, \frac{1}{3})$ or more simply $(\frac{a}{a+1}, \frac{1}{a+1})$. PDFs of this kind can more easily be calculated once this is known via it's area $\frac{b\cdot h}{2}$ or more simply $\frac{b\cdot(\frac{1}{a+1})}{2}$.

Using your example when $P(X > 6Y)$, we proceed with the following:

  1. Find the point of intersection: $\left(\frac{6}{7}, \frac{1}{7}\right)$
  2. Calculate the area: $\frac{1\cdot \frac{1}{7}}{2}=\frac{1}{14}$ and then multiply by f(x,y) to get the answer.

Alternatively, we know that we'll always have 2 triangles to integrate over as given in the image above.

Just following on with the fact that $P(X+Y<1)$ we split the block of the sample space into two, and when $P(X > aY)$ we're always integrating beneath the intersected point. One for $x_1 = (0, \frac{a}{a+1})$ , and $x_2 = \left(\frac{a}{a+1}, 1 \right)$, as for y we get $y_1 = (0, 1-\frac{X}{a})$ and $y_2 = \left(0, 1-X \right)$. It may not work in all cases, perhaps only when $f(x,y) = c$ where $c$ is a constant.

Stackbeans
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