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Probability one random variable is less than another random variable but higher than the same other random variable with a factor

Mathematics Asked by sottoj on December 10, 2021

Consider the two independent random variables $X$ and $Y$. Assume both are uniformly distributed on $[0,1]$. I want to calculate the probability that $X$ lies between some "low" linear transformation of $Y$, say $delta_1Y$, and some "higher" linear transformation of $Y$, say $delta_2Y$, where $delta_1<delta_2Leftrightarrowdelta_1Y<delta_2Y$. Further we know that $delta_1>0$ and $delta_2geq1$. Thus, the probability I want to calculate is $mathbb{P}(delta_1Y<X<delta_2Y)$.

I know that the probability that $X$ is less than $Y$ can be calculated by:

$$mathbb{P}[X<Y]=int_{y=0}^{1}int_{x=0}^{y}P_{X}(x)P_{Y}(y)dxdy$$

where $P_X(x)$ is the PDF of $X$ and $P_Y(y)$ is the PDF of $Y$. A simple solution would be if you simply could calculate the probability as:

$$mathbb{P}[delta_{1}Y<X<delta_{2}Y]=mathbb{P}[delta_{1}Y<X]cdotmathbb{P}[X<delta_{2}Y]$$

But since this implicitly would assume that the two probabilities are independent, I’m pretty sure this is not a viable option. Does anyone have an idea how to solve this?

One Answer

You could start with:

$$begin{aligned}Pleft(delta_{1}Y<X<delta_{2}Yright) & =mathbb{E}mathbf{1}_{delta_{1}Y<X<delta_{2}Y}\ & =int_{0}^{1}int_{0}^{1}mathbf{1}_{delta_{1}y<x<delta_{2}y}dxdy\ & =int_{0}^{1}int_{maxleft(0,delta_{1}yright)}^{minleft(1,delta_{2}yright)}dxdy\ & =int_{0}^{1}minleft(1,delta_{2}yright)-maxleft(0,delta_{1}yright)dy end{aligned} $$

If you already know that $delta_1geq0$ then $max(0,delta_1y)=delta_1y$ and things become a bit easyer.

Go on with discerning cases.

Answered by drhab on December 10, 2021

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