# Probability: pair of random variables

• ashah99
In summary, The conversation is about solving a problem involving integration and probability. The group discusses the correct setup for part (d) and provides pointers for the correct implementation. They also confirm the correctness of the answers for parts (a)-(c).
ashah99
Homework Statement
Problem shown below. Various topics related to the joint pdf and RVs
Relevant Equations
Equations used are shown in working attempt below.
Hello all, I would like to check my understanding and get some assistance with last part of the following question, please.
For part (d), would I use f(x | y) = f(x, y) / f(y) ?

Problem statement:

My attempt at a solution, not too confident in my set-up for part (d). I drew a sketch of the region of integration.

The denominator in the last line must be a double integral, otherwise you can't get a value for it. You need to integrate over the rectangle ##[0,1]\times [0,0.5]##.
You need to break the numerator into two separate integrals: one covering the square on the left and one covering the triangle on the right.

One does not write, as in your central expression on the last line:
$$\frac{f_{XY}(X+Y\leq 1, Y\leq 0.5)} {f_Y(Y\leq 0.5)}$$
You should write something like
$$\frac{Pr(X+Y\leq 1\wedge Y\leq 0.5)} {Pr(Y\leq 0.5)}$$
where the ##Pr## indicates 'probability of' and ##\wedge## means 'and'.

andrewkirk said:

The denominator in the last line must be a double integral, otherwise you can't get a value for it. You need to integrate over the rectangle ##[0,1]\times [0,0.5]##.
You need to break the numerator into two separate integrals: one covering the square on the left and one covering the triangle on the right.

One does not write, as in your central expression on the last line:
$$\frac{f_{XY}(X+Y\leq 1, Y\leq 0.5)} {f_Y(Y\leq 0.5)}$$
You should write something like
$$\frac{Pr(X+Y\leq 1\wedge Y\leq 0.5)} {Pr(Y\leq 0.5)}$$
where the ##Pr## indicates 'probability of' and ##\wedge## means 'and'.

Thank you for the pointers. I very well could still be off on the integral bounds, but if I understood your reply correctly, I get an answer of 0.5875.

That's right. Note that you could have written the numerator as
$$\int_0^{1/2}\int_0^{1-y} f(x,y)\,dx\,dy.$$ Your original attempt just had the limits wrong. What @andrewkirk probably had in mind was
$$\int_0^{1/2}\int_0^{1/2} f(x,y)\,dy\,dx + \int_{1/2}^1 \int_0^{1-x} f(x,y)\,dy\,dx$$ The first integral is the square, and the second integral is the triangular region.

vela said:
That's right. Note that you could have written the numerator as
$$\int_0^{1/2}\int_0^{1-y} f(x,y)\,dx\,dy.$$ Your original attempt just had the limits wrong. What @andrewkirk probably had in mind was
$$\int_0^{1/2}\int_0^{1/2} f(x,y)\,dy\,dx + \int_{1/2}^1 \int_0^{1-x} f(x,y)\,dy\,dx$$ The first integral is the square, and the second integral is the triangular region.

Thank you for your explanation and the integral bounds that you proposed are more intuitive. Do you agree with the final answer for (d)? Also, are my answers for parts (a) - (c) correct? I realize that I have to split the constant c=24 between FX(x) and fY(y) such that both would integrate to 1, i.e. fX (x) = 12x^2(1-x) and fY(y) = 2y.

I got the same answer for (d). The rest looks good as well.

## 1. What is a pair of random variables?

A pair of random variables is a set of two variables that are related to each other in some way, and their values are determined by chance or probability. They are often used in statistical analyses to study the relationship between two variables.

## 2. How are the probabilities of a pair of random variables calculated?

The probabilities of a pair of random variables are calculated using a joint probability distribution, which takes into account the probabilities of each individual variable and their relationship to each other. This can be represented using a table or a graph.

## 3. What is the difference between independent and dependent pair of random variables?

Independent pair of random variables are those that do not affect each other's probabilities, meaning the probability of one variable occurring does not impact the probability of the other variable occurring. Dependent pair of random variables, on the other hand, have probabilities that are influenced by the other variable.

## 4. How can I use probability to predict the outcome of a pair of random variables?

Probability can be used to predict the outcome of a pair of random variables by calculating the joint probability distribution and using it to determine the likelihood of a specific combination of values occurring. However, it is important to note that probability can only provide a prediction and does not guarantee a specific outcome.

## 5. Can a pair of random variables have a negative probability?

No, a pair of random variables cannot have a negative probability. The probability of an event occurring is always between 0 and 1, where 0 represents impossibility and 1 represents certainty. If a calculated probability is negative, it is most likely an error in the calculation.

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