@RemusStanescu Question 2) was answered quite intuitively but incorrectly by Freddy (he'd be right if he focused on conditional expectations rather than probabilities: indeed, P(s>110) < P(s<90) assuming lognormal dynamics for the underlying stock.) This follows from its negative skewness, which is key to your question. First note that call and put prices are nothing more than conditional expectations of the stock price against the respective payoffs (this follows from their being contingent claims.) The negative skewness of the lognormal distribution, which has mean greater than its median, is such that in probabilistic terms, since the average outcome of the stock price at expiry has a less than 1/2 chance of being realized, the only way for calls to be worth more than puts for equally out of the money strikes is for the conditional expectations to weight higher values that offset the loss from probability. And that is just what occurs with the Lognormal. Thus one can say that conditional on being in the money, calls are always worth more than puts. In visual terms consider the longer right tail of the lognormal distribution, which goes off to infinity (whereas put payoffs are bounded by 0) so that higher values despite being less likely in symmetric terms, compensate the final expectation to the loss in probability mass. Remember, pricing is usually nothing more than integrating a payoff over a probability distribution of some dimension. see here:http://en.wikipedia.org/wiki/File:Comparison_mean_median_mode.svg
SRKX does enough to answer your ATM question. see here for more: http://www.ederman.com/new/docs/qf-Illusions-dynamic.pdf
In terms of the above, ATM puts = ATM calls when S=K (r=q=0) because there is a correction of 1/2*{sigma^2} to the drift of the stock dynamics that offsets the probability that P(S>K)<.5 when S is lognormally distributed. But this is a model dependent reason: just read the paper in the link above for the stronger, model independent reason.
On edit (to place all comments in one place:)
I did not purge anything. Looks like the moderator did, and threw away all of the formulas i used to prove that if we remain in the risk neutral world, one cannot make the approximation *(1)* $S(t) \exp^{r-\frac{\sigma^2}{2}} \sim S(t)$ without introducing arbitrage. All the probabilities in this question require the risk neutral measure which allows for the correct integration of $SN\left(d_1\right)-Ke^{-r(T-t)}N\left(d_2\right)$ where
$d_1=\frac{\ln(S/K)+\left(r+\frac{1}{2}\sigma^2\right)(T-t)}{\sigma \sqrt{T-t}}$,
$d_2=d_1-\sigma \sqrt{T-t}$ and $N(x)=\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^x e^{-\frac{1}{2}s^2}ds$ (assuming no divs and the formula for the call price, with no loss in generality.) Notice $d_1, d_2$ are limits of integration. If one makes the approximation *(1)*, the integral evaluates so as to introduce arbitrage into the BS formula, in particular, PC parity is violated. Again, in simplest terms, just look at $N(d_2)$, which is the risk neutral probability that S>k at expiry. Plug in the numbers and you'll see proof positive that in the BS framework, it is not the unconditional probability of being in the money that makes calls more expensive versus puts for symmetrically OTM strikes (in a world where there is no implied volatility skew, i.e., each strike gets the same vol), it is the expectation conditional on the terminal stock price being in the money that makes it so. Of course i've lost my breath trying to make you see this, to actually prove it to yourself, to no avail. You simply aren't thinking closely about what I'm saying. And for the love of 'i have a life', I ain't copying anything from anywhere. Why would I care to do that? I am here because I noticed your reference to the unconditional probability of being in the money is wrong in the risk neutral world, and simply found it interesting to flesh this out. In case you're wondering, my intuition was guided simply by checking the claim with the good ole, $N(d_2)$.
I actually pasted one of my purged comments in a notepad that is still open, so let me offer this up for further proof:The right comparison is between: $P(S(T)> 110) \sim 1/2 - P(median<S(T)<110)$ & $P(S(T)<90) \sim 1/2-P(90<S(T)<median)$ As soon as vol>0, the median is less than the mean which means there is some substantial probability mass getting wedged between 100, the mean, and the median. If one then accounts for this, the following becomes true: $P(median<S(T)<110)$ > $P(90<S(T)<median)$, and thus $P(S(T)<90)$ > $P(S(T)>110)$,
and therefore we are back to step 1, trying to understand why the call is worth more than the put. It is then the reasoning based on expectation that provides the answer.
