Expected Value of Exponential with IID Uniform Argument

Let us assume that we generate a discrete random variable $I \in \{-1,1\}$.

Then, we know that $E\{I\} = \frac{1}{2} \cdot 1 - \frac{1}{2} \cdot 1 = 0$.

However, what is $E\{e^{jI}\}$?

Where $j$ is the imaginary unit and $e$ is the exponential. Can we apply the same logic and write:

$E\{e^{jI}\} = \frac{1}{2} e^{j} + \frac{1}{2} e^{-j} = cos(1)$

I know that linearity doesn't apply, but since we know that I is uniform, can we just say that:

$E\{e^{jI}\} = e^{j\cdot0} = e^{0} = 1$

Which one is correct?

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I don't know what you mean by saying that $I$ is IID. There is only one random variable here, so there is nothing for it to be independent from or identically distributed to. – Trevor Wilson Feb 28 '13 at 1:24
Sorry, its part of a larger question where I comes from an IID process. – TheDude Feb 28 '13 at 1:26

The first argument is correct. As you suspected the assumption $E[f(I)]=f(E[I])$ is not always valid. In fact the only functions $f$ for which this equality holds for all random variables $I$ are the affine functions (linear functions plus a constant) and this does not include the exponential function.
For a more obvious counterexample consider the absolute value function: $E[|I|] = 1$ but $|E[I]| = 0$.