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# Proof, modular arithmetic, probs & stats Watch

1. (Original post by SubAtomic)
had to look at evaluating integrals with infinity, so I get this

Is this right? If so I think I may just about get it
Well have you understood why M(t) is always positive now?
2. (Original post by Smaug123)
Pretty much, yep - the infinite integral should technically be evaluated as a limit, but this works fine if you just use lim(e^-x) = 0 as x->inf.
Yeah I did all the lim stuff here but didn't want to tex it so that is what I expected, so

3. (Original post by shamika)
Well have you understood why M(t) is always positive now?
Possibly, because the condition t < lambda means if lamda negative then Mx(t) positive and if lambda positive t < lambda means Mx(t) positive ?

Is that just waffle? Yes because it is given that lambda > 0 hahaha

Am I a lost cause
4. (Original post by SubAtomic)
Possibly, because the condition t < lambda means if lamda negative then Mx(t) positive and if lambda positive t < lambda means Mx(t) positive ?

Is that just waffle? Yes because it is given that lambda > 0 hahaha
Pretty much waffle; isn't lambda restricted to be positive?

OK lets go back to the integral you've evaluated. Why have you got rather than ?

And once you've understood that, tell me what happens to as x becomes very large.

Finally, putting all of that together you'll get the expression on the page you've scanned.
5. (Original post by shamika)
Pretty much waffle; isn't lambda restricted to be positive?
Lol, thought so.

(Original post by shamika)
OK lets go back to the integral you've evaluated. Why have you got rather than
Because t < lambda so F(b) has the e^{- \infty} in the numerator.

(Original post by shamika)
And once you've understood that, tell me what happens to as x becomes very large.
Tends to 0

(Original post by shamika)
Finally, putting all of that together you'll get the expression on the page you've scanned.
Yep
6. Many thanks to you all

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