Jointly continuous random variables

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SUMMARY

The discussion centers on calculating the probability that the reimbursement for random losses X and Y, defined by the joint density function f(x,y) = e^-(x + y) for x > 0 and y > 0, is less than 1. The initial attempt at solving the double integral p(X + Y < 1) led to confusion regarding the limits of integration, which were incorrectly set from 0 to infinity instead of 0 to 1. The correct approach involves reevaluating the inner integral, which simplifies to e^{-x} ∫_0^{1-x} e^{-y} dy.

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Homework Statement



Let X and Y be random losses with joint density function

f(x,y) = e^-(x + y) for x > 0 and y > 0 and 0 elsewhere

An insurance policy is written to reimburse X + Y:
Calculate the probability that the reimbursement is less than 1.

Homework Equations



Have not learned independence for jointly cont r.v's yet



The Attempt at a Solution



p(X + Y < 1) = p(Y < 1 - X) = ##\int_{0}^{\infty}\int_{0}^{1-x} e^{-(x+y)} dydx##

When I go through solving this double integral I get the following

##-e^{-x} - xe^{-1}## evaluated from 0 to ∞.

However as x → ∞ the above function diverges. Maybe I calculated the integral wrong, I have done it over and over, and cannot seem to find where it could be wrong.
 
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Never mind I figured out my mistake. The limits of the first integral should not be 0 to infinity. They are 0 to 1
 
DotKite said:
Never mind I figured out my mistake. The limits of the first integral should not be 0 to infinity. They are 0 to 1

That was not your only mistake: you need to re-do the inner integral
[tex]\int_0^{1-x} e^{-(x+y)} \, dy = e^{-x} \int_0^{1-x} e^{-y} \, dy.[/tex]

Edit: Oh... maybe you wrote the indefinite x- integral of inner y-integral; in that case, you are correct.
 
Last edited:

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