Is P(x + a > y + b) Always 0.5 for Independent Variables?

[Schlotkins]

Good evening:

I have a probability proof question that is driving me crazy. I feel
like I must have forgot an easy trick. Any help is GREATLY
appreciated. Here's the setup:

Let's assume a,b are indepedent random variables from cummulative
distribution F.


I think it's safe to say:


P( a > b) = .5


Now, let's assume x,y are independent random variables from CDF G.
Again:


P(x > y) = 0.5


Assume CDFs G and F are indepedent. Now it seems straightforward that:


P(x + a > y + b) = 0.5


but I don't know how to show it without assuming a distribution type.


Again, any help is appreciated.


Thank you,
Chris

 

[mathman]

If you are familiar with characteristic functions it is simple.
Let f and g be the characteristic functions of F and G.
Then S= x - y + a - b will have a ch. fcn. f(t)f(-t)g(t)g(-t).
This means that S has a symmetric distribution.

 

[Schlotkins]

Thank you for the tip and the response - I think I have it solved. On an aside, that trick wouldn't work for P(ax < by) right? It seems obvious that the P(ax - by < 0 ) = P (by - ax < 0)= .5, but of course characteristic functions are most useful for sums.

Thanks again for your assistance.
Chris