Finding the probability of $P(40.5<Y<48.9|X>=68.6)$

Question

Two students go to a pizza place every week. Let $X$ and $Y$ be the weekly spend of each student at the pizza place. Assume that $X$ and $Y$ have a bivariate normal distribution with

$\mu_X=60.6, \sigma_X=11.2, \mu_Y=46.8, \sigma_Y=8.4, \rho=0.94$

a) Find $P(40.5<Y<48.9|X\geq 68.6)$

EDIT: I think I have a solution. Can someone please verify?

$$f_{XY}(x,y)=\frac{1}{2\pi(11.2)(8.4)\sqrt{1-(0.94)^2}}*\exp{-\frac{1}{2(1-(0.94)^2)}[(\frac{x-60.6}{11.2})^2+(\frac{y-46.8}{8.4})^2]-2(0.94)\frac{(x-60.6)(y-46.8)}{(11.2)(8.4)}}$$

$P(40.5<Y<48.9|X\geq 68.6)=\frac{P(40.5<Y<48.9, X\geq 68.6)}{P(X\geq 68.6)}$

$P(X\geq 68.6)=P(Z\geq 0.7143)=0.23752$

$$\int_{40.5}^{48.9}\int_{68.6}^{\infty}\frac{1}{2\pi(11.2)(8.4)\sqrt{1-(0.94)^2}}*\exp{-\frac{1}{2(1-(0.94)^2)}[(\frac{x-60.6}{11.2})^2+(\frac{y-46.8}{8.4})^2]-2(0.94)\frac{(x-60.6)(y-46.8)}{(11.2)(8.4)}} dx dy = 0.0051345$$

$P(40.5<Y<48.9|X\geq 68.6) = \frac{0.0051345}{0.23752}$

b) Find $P(40.5<Y<48.9)$

$=P(-0.75<Y<0.25)=0.3721$

c) Find $P(40.5<Y<48|X=68.6)$

$E(Y|X)=46.8+(0.94)(\frac{8.4}{11.2})(68.6-60.6)=52.44$

$Var(Y|X)=(8.4)^2-(0.94)^2(8.4)^2=8.21$

$P(-4.17<Z<-1.24)=0.1075$

d) What is the relationship between these probabilities?

The relationship is that part (a) is smaller than the other two probabilities since you are dividing out a larger number. Also part(b) is univariate while part (a) and part (c) is bivariate.

I am mostly just trying to see whether my solutions for parts (a) and part (d) are correct because those are the ones I am most confused about.

Answer 1

Parts (a), (b), (c): Computations

For the computational parts (a), (b), and (c) I checked your math and I got the same results, including the intermediate results. Nice. I used Mathematica for these (code at the bottom) and I'm guessing you used some totally different software, but that's a good thing - it just means it's all the more reassuring when our answers agree :)

Part (d): Interpretations

This is the tricky part. Problem (d) asks us to understand the relationship between the different probabilities we've been computing, so let me write them out. I think all the numbers can get distracting, so I'm going to use $I = (40.5, 48.9)$ and $r = 68.6$. The computed probabilities are:

• A: $P(Y \in I | X \ge r) \approx 0.02$
• B: $P(Y \in I) \approx 0.37$
• C: $P(Y \in I | X = r) \approx 0.11$

All of these are talking about the chances that $Y$ lies in the interval $I$. Statement B is the simplest: it tells us the overall probability that $Y \in I$ is pretty high (37%).

A is next. In this problem we kept $r$ constant, but think about what would happen if we changed $r$ and repeated the same calculations. What would happen if we made $r$ really small? Answer: we'll have $$P(Y \in I : X \ge r) \to P(Y \in I) \text{ as }r \to -\infty.$$ (Do you see why?)

Last is statement C. In part C, we used a 1D integral to do computations with one specific value of $X$. You can think of fixing $X$ as "choosing a 1D cross-section of the 2D probability space". If you add up the results from lots of these cross-sections, you can find the answers to some 2D questions. That intuition can be written in terms of integrals: we can integrate the area under lots of 1D curves to get the volume under a 2D surface. In particular, $$ \begin{align} P(Y \in I) &= \int_{-\infty}^\infty P(Y \in I : X = t) dt \\ P(Y \in I : x \ge r) &= \int_r^\infty P(Y \in I : X = t) dt \\ \frac{d}{dr} P(Y \in I : x \ge r) &= P(Y \in I : X = r) \end{align} $$

Mathematica code

dist = BinormalDistribution[{60.6, 46.8}, {11.2, 8.4}, 0.94];

solve[xmin_, xmax_, ymin_, ymax_] := 
 NIntegrate[PDF[dist, {x, y}], {x, xmin, xmax}, {y, ymin, ymax}, 
  WorkingPrecision -> 15]

(* Here are the function calls I used in part A: *)
solve[68.6, Infinity, 40.5, 48.9]
solve[68.6, Infinity, -Infinity, Infinity]

(* ...and so on for the other parts. *)