We know that P(X = 1) = 3/4, and
P(X = 2.10^-n) = 5^-n
, n = 1, 2, 3, . . . .
Calculate the exact value of FX(0.001).

Question

We know that P(X = 1) = 3/4, and
 P(X = 2.10^-n) = 5^-n
, n = 1, 2, 3, . . . .
Calculate the exact value of FX(0.001).

miracrown · Answer

Do we know anything else about P?

anonymous · Answer

no just this info

miracrown · Answer

I am thinking ...

anonymous · Answer

What is $FX(0.001)$?

anonymous · Answer

I think you wrote it wrong.

miracrown · Answer

If we don't know anything else about P, I don't think the problem can be done
if there aren't any other restrictions on P, it can do anything in between those values
so it could have any value 	at 0.001

anonymous · Answer

no i wrote it right
i think it means P( X< 0.001)

anonymous · Answer

$$
P(X<0.001) = \int_{-\infty}^{0.001} P(X=x)~dx
$$

miracrown · Answer

Oh, so those are probabilities? That makes more sense
	
So, we can do this by either summing the infinite series
or, find the values for x>= 0.001 and subtracting that probability from 1

anonymous · Answer

Wait, if it is a discrete probability...

anonymous · Answer

i guess this does not deal with integral, i'm agree with Miracrown ( summing the infinite series )

anonymous · Answer

n=1,2,3,...

anonymous · Answer

I think at $n=4$, we have gone too far.

miracrown · Answer

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