Probability that sum of two random variables is greater than 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?

 

[mfb]

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.

 

[Ray Vickson]

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

 

[D H]

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.

 

[rayge]

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.