Math question

[Borodog]

Let x be a random number, normally distributed around x1 with standard deviation s1, and x' be a second random number, normally distributed around x2 with standard deviation s2.

What is the probability that x' is greater than x?

[hmkpoker]

On first glance it seems like 50-50, unless I'm missing something.

[Borodog]

[ QUOTE ]
On first glance it seems like 50-50, unless I'm missing something.

[/ QUOTE ]

You are missing something. Let x1 = 0 and s1 = 1, and x2 = 1000 and s2 = 1.

Still think it's 50-50 that x1 is greater than x?

[kyro]

[ QUOTE ]
On first glance it seems like 50-50, unless I'm missing something.

[/ QUOTE ]

He's asking for a formula in terms of the variables he's given.

I should know this off the top of my head as I'm tutoring a girl in it right now, but I'm blanking big time. Something like...

(x2-x1) - (something)
_____________________
s2*s1


That's not correct, but I think I'm on the right track. Oh, of course you need to use z-scores and the table.

[hmkpoker]

[ QUOTE ]
[ QUOTE ]
On first glance it seems like 50-50, unless I'm missing something.

[/ QUOTE ]

You are missing something. Let x1 = 0 and s1 = 1, and x2 = 1000 and s2 = 1.

Still think it's 50-50 that x1 is greater than x?

[/ QUOTE ]

Give me a break, that's results-oriented thinking. You said that x1 and x2 were random numbers, it was just as likely that x1 could have been 1000 and x2 could have been 0.

[Borodog]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
On first glance it seems like 50-50, unless I'm missing something.

[/ QUOTE ]

You are missing something. Let x1 = 0 and s1 = 1, and x2 = 1000 and s2 = 1.

Still think it's 50-50 that x1 is greater than x?

[/ QUOTE ]

Give me a break, that's results-oriented thinking. You said that x1 and x2 were random numbers, it was just as likely that x1 could have been 1000 and x2 could have been 0.

[/ QUOTE ]

Uh, no. I said x and x' are random numbers, not x1 and x2.

[hmkpoker]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
On first glance it seems like 50-50, unless I'm missing something.

[/ QUOTE ]

You are missing something. Let x1 = 0 and s1 = 1, and x2 = 1000 and s2 = 1.

Still think it's 50-50 that x1 is greater than x?

[/ QUOTE ]

Give me a break, that's results-oriented thinking. You said that x1 and x2 were random numbers, it was just as likely that x1 could have been 1000 and x2 could have been 0.

[/ QUOTE ]

Uh, no. I said x and x' are random numbers, not x1 and x2.

[/ QUOTE ]

But you didn't provide the values of x1 and x2, or give any information as to what they might be. If we are not to know what they are, they are for all practical purposes random as well.

[Borodog]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
On first glance it seems like 50-50, unless I'm missing something.

[/ QUOTE ]

You are missing something. Let x1 = 0 and s1 = 1, and x2 = 1000 and s2 = 1.

Still think it's 50-50 that x1 is greater than x?

[/ QUOTE ]

Give me a break, that's results-oriented thinking. You said that x1 and x2 were random numbers, it was just as likely that x1 could have been 1000 and x2 could have been 0.

[/ QUOTE ]

Uh, no. I said x and x' are random numbers, not x1 and x2.

[/ QUOTE ]

But you didn't provide the values of x1 and x2, or give any information as to what they might be. If we are not to know what they are, they are for all practical purposes random as well.

[/ QUOTE ]

Uh, no? They're just parameters. If I ask you for the kinetic energy of an object of mass m and speed v, you don't get to assume m and v are random.

I'm saying, if I give you two distributions of given (but different) mean and standard deviation, what is the probability that a random deviate from one distribution will be greater than a random deviate from the other. Obviously I don't want a number, which you can't produce without the particular means and standard deviations, but rather an expression involving those parameters.

[hmkpoker]

ahhhhhhh

[jason1990]

[ QUOTE ]
Let x be a random number, normally distributed around x1 with standard deviation s1, and x' be a second random number, normally distributed around x2 with standard deviation s2.

What is the probability that x' is greater than x?

[/ QUOTE ]
If x and x' are jointly normal, then x-x' is normal. Its mean is m=x1-x2 and its variance is s^2=s1^2+s2^2-2c, where c is the covariance of x and x'. Hence, x-x'=sN+m, where N is a standard normal and

P(x' > x) = P(sN + m < 0) = P(N < -m/s),

which you can look up in a table.

[Borodog]

What is the "covariance" of x and x'?

Edit: What I mean is, isn't it zero here?

[gumpzilla]

At least link them to the horse racing question next time.

[bigpooch]

Cov (x,x') = E[x*x']-x1*x2.

If x and x' are independent, then this is zero because
then E[x*x'] = E[x]*E[x'] = x1*x2.

On the other hand, x and x' need not be independent.