> (value of put spread) / (distance between strikes) ~ P(stock expiring below midpoint of strikes)
While the pic/chart in this post containing discrete probabilities/weighted payoffs is legit, I was thinking about this more yesterday and believe there's an easier way to explain it.
x * (width of strikes) + y * (avg payoff when stock is between strikes) = (value of put spread)
=> x * (width of strikes) + y * ((width of strikes) / 2) = (value of put spread)
=> x + (y / 2) = (value of put spread) / (width of strikes)
Assuming probability stock can land between strikes is equally distributed.. (y/2) happens to be probability stock finishes below mid point. (Within the strikes)
x + y/2 = cumulative probability stock finishes below mid point of strike.
I believe because above assumption is almost true and not quite, is why the equation also is approximately true.
Awesome post.. thanks for connecting the dots for me!
> (value of put spread) / (distance between strikes) ~ P(stock expiring below midpoint of strikes)
While the pic/chart in this post containing discrete probabilities/weighted payoffs is legit, I was thinking about this more yesterday and believe there's an easier way to explain it.
(I didn't fully follow it/understand how this conclusion came about when I first read it at https://www.globalcapital.com/article/28mwrr0lap7g5gyuamznl/derivatives/option-prices-imply-a-probability-distribution)
Building off of part 1 of ur post:
P(stock expires below short strike) = x
P(stock expires between strikes) = y
x * (width of strikes) + y * (avg payoff when stock is between strikes) = (value of put spread)
=> x * (width of strikes) + y * ((width of strikes) / 2) = (value of put spread)
=> x + (y / 2) = (value of put spread) / (width of strikes)
Assuming probability stock can land between strikes is equally distributed.. (y/2) happens to be probability stock finishes below mid point. (Within the strikes)
x + y/2 = cumulative probability stock finishes below mid point of strike.
I believe because above assumption is almost true and not quite, is why the equation also is approximately true.
Awesome post.. thanks for connecting the dots for me!
That sounds like the derivation I forgot thank you! Awesome work.
I appreciate that you thought about this post hard!
On the following statement:
If the 100/85 put spread is $5.17 and its maximum value is $15, then if you buy it, you are getting 9.83 to 5.17 odds or 2.9 to 1.
Would that be 2.9 to 1 or 1.9 to 1?
1.9...great catch!