Sum of random variables

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womp3

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So the question states:

"Let X and Y be independent random variables, each exponentially distributed with parameter lambda. Find the density function and the distribution function of X + Y ."

I mostly understand what density and distribution functions are but I never learned how to calculate them from such little information. The only hint that was given said that "the probability density function of the sum of two independent random variables X and Y, with the respective probability density functions fx and fy, is given by the formula: fx+y(t) = integral from -infinity to +infinity of fx(s)*fy(t-s)*ds"

Any ideas would be very appreciated.
 
womp3 said:
So the question states:

"Let X and Y be independent random variables, each exponentially distributed with parameter lambda. Find the density function and the distribution function of X + Y ."

I mostly understand what density and distribution functions are but I never learned how to calculate them from such little information. The only hint that was given said that "the probability density function of the sum of two independent random variables X and Y, with the respective probability density functions fx and fy, is given by the formula: fx+y(t) = integral from -infinity to +infinity of fx(s)*fy(t-s)*ds"

Any ideas would be very appreciated.
Click to expand...

google convolution
 
OK. I used convolution for the density function. Do I also use that for the distribution function or just add the functions normally?
 
womp3 said:
OK. I used convolution for the density function. Do I also use that for the distribution function or just add the functions normally?
Click to expand...

I'm not sure I'm understanding what you think the difference between density functions and distributions is? A continuous distribution function is a probability density function.

If X is distributed as pX(s) and Y as pY(s) then if Z = X + Y, then Z will have the distribution function given by pX(s) convolved with pY(s)

This by the way is the formula you mentioned in your original post.
 
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