# Finding the probability a set of random values are less than a specified value

If $$X$$ and $$Y$$ are independent random variable following the normal distribution $$N(\mu, \sigma^2)$$ with mean $$\mu$$ and variance $$\sigma^2$$ such that $$X \sim N(\mu_{X}, \sigma_{X}^2)$$ and $$Y \sim N(\mu_{Y}, \sigma_{Y}^2)$$, and if $$T$$ is some constant, then the probability $$P(X-Y

$$P(Z

where $$Z =X-Y$$, $$\mu_{Z}=\mu_{X}-\mu_{Y}=$$, and $$\sigma_{Z}^2=\sigma_{X}^2 + \sigma_{Y}^2$$. (The detailed explanations can be found here.)

I want to extend this to the case of random vectors $$\mathbf{X}$$ and $$\mathbf{Y}$$. Let us consider the random vectors of length $$L=3$$.

Then, for $$X_1$$ conditioned on the $$\mathbf{Y}$$

$$P(X_1-Y_1 $$P(X_1-Y_2 $$P(X_1-Y_3

and for $$X_2$$ conditioned on the $$\mathbf{Y}$$ $$P(X_2-Y_1 $$P(X_2-Y_2 $$P(X_2-Y_3

and for $$X_3$$ conditioned on the $$\mathbf{Y}$$ $$P(X_3-Y_1 $$P(X_3-Y_2 $$P(X_3-Y_3

My question is how do I obtain the total probability that $$Z$$ is less than $$T$$.

Would it be $$P(\mathbf{X}-\mathbf{Y}?

Appreciate any help and advice.

• I can't follow what you mean by "average probability". – gunes Aug 14 '20 at 11:04
• @gunes Sorry, I meant the total probability . – nashynash Aug 14 '20 at 11:09
• If it's not the equation in the end that you want, what does total probability mean? Do you mean $$P(X-Y<T):=P(X_i-Y_j<T)$$ for all $i,j$? i.e. $$P(X_1-Y_1<T,X_1-Y_2<T,...,X_3-Y_3<T)$$ all at the same time. – gunes Aug 14 '20 at 11:24
• ... or do you mean $P(X-Y<T) := P(X_1-Y_1<T, X_2-Y_2<T, X_3-Y_3<T)$? – jan-glx Aug 14 '20 at 11:32
• @gunes Yes, I think what you describe is what I am looking for. How do I go about formulating the right side? – nashynash Aug 14 '20 at 12:23

## 1 Answer

If the question is about the following probability $$\varrho=\mathbb P(X_1−Y_1 it is given by$$\varrho=\mathbb P(\max_{1\le i,j\le 3}(X_i−Y_j)≤T)$$ since the event that the pairwise differences all are less than $$T$$ is equivalent to the largest pairwise difference being less than $$T$$. Furthermore, the largest difference $$X_i-Y_j$$ is equal to the difference between the largest $$X_i$$, $$X_{(3)}$$, and the smallest $$Y_i$$, $$Y_{(1)}$$ hence$$\varrho=\mathbb P(\max_{1\le i\le 3}(X_i)−\min_{1\le j\le 3}(Y_j)≤T)$$ The density of $$X_{(3)}$$ is \begin{align*}f(z)&=\Phi(1/\sigma^X_1(z-\mu^X_1))\Phi(1/\sigma^X_2(z-\mu^X_2))\varphi(1/\sigma^X_3(z-\mu^X_3))+ \cdots\\ &+ \Phi(1/\sigma^X_3(z-\mu^X_3))\Phi(1/\sigma^X_2(z-\mu^X_2))\varphi(1/\sigma^X_1(z-\mu^X_1))\end{align*} and the density of $$Y_{(1)}$$ is \begin{align*}g(w)&=\Phi(-1/\sigma^Y_1(w-\mu^Y_1))\Phi(-1/\sigma^Y_2(w-\mu^Y_2))\varphi(1/\sigma^Y_3(w-\mu^Y_3))+ \cdots \\&+ \Phi(-1/\sigma^Y_3(w-\mu^Y_3))\Phi(-1/\sigma^Y_2(w-\mu^Y_2))\varphi(1/\sigma^Y_1(w-\mu^Y_1))\end{align*} where $$\varphi$$ is the $$\mathcal N(0,1)$$ pdf and $$\Phi$$ is the $$\mathcal N(0,1)$$ cdf, respectively. The pdf of $$X_{(3)}-Y_{(1)}$$ follows as a convolution integral but cannot be expressed in closed form.

• Is it to be understood that (a) $\varphi(\cdot)=1-\Phi(\cdot)$; (b) $z=X$; and (c) $w=Y$? If I am mistaken, could you kindly clarify? – nashynash Aug 14 '20 at 15:28
• Thank you for the clarification. Could we apply this analysis to the case where we have the i.i.d. RVs $A,B,...,J$, where $A\sim\mathcal N(\mu_A,\sigma^2_A),..J\sim\mathcal N(\mu_Z,\sigma^2_Z)$? Then, the total probability that the difference between two consecutive RVs is less than $T$ is $P_{tot}=\mathbb P(B-A<T,C-B<T,...,J-I<T)=\mathbb P(J-A<T)$? If not, I could create a separate page. – nashynash Aug 15 '20 at 4:07
• I repeat, this is not correct. This is basic maths, not probability: if $B-A<T$ and $C-B<T$, this is not the same thing as $C-A<T$. Make a try with a few values of $A,B,C$. – Xi'an Aug 16 '20 at 5:42
• Okay!.Thank you very much, Xi'an for your time and explanations. Truly appreciate it. I will work out the math. – nashynash Aug 16 '20 at 5:45