# Probability that sum of two random variables is greater than 1

• rayge
In summary: I'll try to be more careful in the future when setting up my integrals.In summary, the problem asks for the probability that the sum of two random variables, X_1 and X_2, is greater than or equal to 1. Using the joint pdf of 1/x_1, the correct integral to evaluate is \int_0^1\int_{1-x_1}^1 1/x_1 dx_2 dx_1. This integral evaluates to 1 and represents the portion of the sample space where x_1+x_2≥1. Another approach is to use a conditioning argument, where the probability is found by conditioning on the value of x_1.
rayge

## Homework Statement

Let us choose at random a point from the interval (0,1) and let the random variable $X_1$ be equal to the number which corresponds to that point. Then choose a point at random from the interval (0,$x_1$), where $x_1$ is the experimental value of $X_1$; and let the random variable $X_2$ be equal to the number which corresponds to this point.

Compute $P(X_1 + X_2 >= 1)$

## Homework Equations

The joint pdf is $1/x_1 , 0<x_2<x_1<1$

## The Attempt at a Solution

Many. For one, set $Y=X_1+X_2$. Then find $P(Y>=1)$. Then evaluate $\int_0^1\int_{1-x_2}^1 1/x_1 dx_1dx_2$. Evaluating this integral gives me zero.

The other solutions I come up with end up giving me $ln(0)$, which is undefined. Any suggestions on how to approach this?

How can you get zero from the integral? 1/x_1 is always positive, the result cannot be zero.
I don't see how you consider the sum in this integral.

for a given x_1, what is the probability that the sum is larger than 1?
With this probability, you can use a one-dimensional integral to get the answer.

rayge said:

## Homework Statement

Let us choose at random a point from the interval (0,1) and let the random variable $X_1$ be equal to the number which corresponds to that point. Then choose a point at random from the interval (0,$x_1$), where $x_1$ is the experimental value of $X_1$; and let the random variable $X_2$ be equal to the number which corresponds to this point.

Compute $P(X_1 + X_2 >= 1)$

## Homework Equations

The joint pdf is $1/x_1 , 0<x_2<x_1<1$

## The Attempt at a Solution

Many. For one, set $Y=X_1+X_2$. Then find $P(Y>=1)$. Then evaluate $\int_0^1\int_{1-x_2}^1 1/x_1 dx_1dx_2$. Evaluating this integral gives me zero.

The other solutions I come up with end up giving me $ln(0)$, which is undefined. Any suggestions on how to approach this?

You could use a conditioning argument, where you condition on ##x_1##. In other words, find ##P(X_1+X_2 \geq 1|X_1 = x_1)##, etc.

Alternatively, you can use the joint density, but you need to apply the correct integration. To do that, first draw a picture to help you get the region of integration correctly

rayge said:

## The Attempt at a Solution

Many. For one, set $Y=X_1+X_2$. Then find $P(Y>=1)$. Then evaluate $\int_0^1\int_{1-x_2}^1 1/x_1 dx_1dx_2$. Evaluating this integral gives me zero.
As has been noted, that integral does not evaluate to zero. (It evaluates to one.)

Much more importantly, that is the wrong integral. Look what happens when x2 is, for example, 3/4. The integration limits for x1 are 1/4 to 1. When x1 is 1/4, the maximum possible value for x2 is 1/4. An x2 value of 3/4 is not possible.

Bottom line: Your integral covers more than the sample space.

If you want to use this approach, I suggest drawing a picture. You want the portion of the sample space for which x1+x2≥1.

Thanks a lot for the help everyone. Once I thought carefully about summing all values of $x_1$, the proper limits of the integrals was a lot more clear.

## 1. What is the meaning of "probability that sum of two random variables is greater than 1"?

The probability that the sum of two random variables is greater than 1 refers to the likelihood that the combined value of two random variables will be greater than 1 when they are added together. This can also be interpreted as the chance that the sum of the two variables will fall within a certain range above 1.

## 2. How is the probability of the sum of two random variables being greater than 1 calculated?

The probability of the sum of two random variables being greater than 1 can be calculated using various methods, depending on the distribution of the variables. For continuous random variables, it can be calculated by integrating the joint probability density function over the specified range. For discrete random variables, it can be calculated by summing the probabilities of all possible outcomes that meet the criteria of being greater than 1.

## 3. What factors affect the probability that the sum of two random variables is greater than 1?

The probability of the sum of two random variables being greater than 1 is affected by the individual probability distributions of the two variables, as well as their correlation. Other factors such as sample size and the range of values for each variable can also impact the probability.

## 4. Can the sum of two random variables ever be guaranteed to be greater than 1?

No, the sum of two random variables can never be guaranteed to be greater than 1. This is because the values of the variables are determined randomly and cannot be controlled. The probability of the sum being greater than 1 can be increased or decreased, but it can never be certain.

## 5. How is the probability of the sum of two random variables being greater than 1 useful in real-life applications?

The probability of the sum of two random variables being greater than 1 is useful in various fields, such as finance, statistics, and machine learning. It can be used to analyze and predict the outcomes of events that involve multiple random factors. For example, it can be used to calculate the probability of a stock portfolio having a certain return, or the likelihood of a machine learning algorithm producing a certain result.

• Calculus and Beyond Homework Help
Replies
7
Views
1K
• Calculus and Beyond Homework Help
Replies
9
Views
779
• Set Theory, Logic, Probability, Statistics
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
6
Views
2K
• Calculus and Beyond Homework Help
Replies
11
Views
1K
• General Math
Replies
3
Views
680
• Calculus and Beyond Homework Help
Replies
9
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
821
• Precalculus Mathematics Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
2K