Probability Theory - Expected Value.

[rad0786]

Hey.. im having some problems with this homework question, i thought perphaps somebody can help me on it.

-- Let X be a random variable with probability mass function
x P(x)
-1 p
1 1 - p

Find the value of the constant C not equal to 1 such that E[c^X] = 1
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So this is my work so far. We have E[X] = 1 -2p and E[c^X] = 1 ...

E[X] = 1 -2p
E[c^X] = 1

then c^X = c^(1-2p) = 1 so...

c^(1-2p) = 1

I do the logarithms....

(1-2p)logc = log1
(1-2p)logc = 0
log c = 0
c = 1

but that is no true, since c cannot equal 1. What am i doing wrong?

The only other solution i can see is 1^(1-2p)^-1 = 1/(1-2p)

[rad0786]

sorry.. the mass funtion looks awkward above... its actually...

x p(x)
-1 p
1 p-1

[HallsofIvy]

You seem to have a completely wrong idea about what's going on here!
No, c^X is not c^(1-2p). That would be cE(x) which is irrelevant.

What your probability function says is that x can have one of only two values: -1 and 1, with probabililty p and 1-p respectively. That means that c^x can have one of only two values: c-1= 1/c and c, with probability p and 1-p respectively. The "expected value" is the sum of the each possible value times its probability: p/c+ c(1-p)= 1. Solve that for c in terms of p.

[rad0786]

Hey... I see the error I made now....
for solving... i managed to simplyfy the equation to 0 = (1-p)c^2 - c + p

This is the same as... 0 = (1-p)x^2 - x + p and the idea is to solve for x or p. Believe me, I tried EVERYYTHING to solve this, even the quadratic. I just keep finding my self at a ded end. I know that c=1 works just fine, but the question says that c cannot equal to one.
Anybody have any ideas?

[rad0786]

Ive been stuck on this for the whole day... how on earth do i solve for c?

(1-p)C^2 - C + p = 0 //or// p/C + C(1-p) = 1

I just keep getting 1 as an answer and the questions says C not equal to 1