Synthetic position

Hi all - I came over the following question in Qbank:

To create a synthetic short position in a stock, an investor can buy:

A) a call option on the stock and sell a put option on the stock.

B) both a call option on the stock and a put option on the stock.

C) a put option on the stock and sell a call option on the stock.

The correct answer is C) - however it looks like for me that C) is the long synthetic position on the stock not the short one.

Could anyone please explain?

Thanks many times,


Short position in stock- gain when stock goes down, if stock moves up, loss

Long put- gain if.stock goes down

Short call- loss if stock moves up.

Combine both and u have position equivalent to short position in stock.

Draw a payoff graph with a long put and short call with the same strike price.

For synthetics, effectively use the Put Call Parity (Stock + Put = Call) and rearrange that formula to work it out. So a short position (I.e. a negative in the Put call parity formula) on the stock becomes Put - Call = - Stock


Is the Xert used anywhere? If yes, in what type of situations?

Don’t get confused, look at it in a logical way

You want to replicate a short position, when you say a synthetic short position.

So , a short position gives you profits from a down move and losses from the stock moving up.

So, if someone had to replicate it, in a way that he profits from down move and loses from up moves. So,

He would buy a put option (he profits from a down move) and sell a call option (he loses when stock goes up).

So the cash flows from the above would be equivalent to a short position in a stock

hi guys, thanks many times, I guess I was writing the put call parity not correctly :slight_smile:

It’s not explicitly referenced anywhere, but if you play through the synthetic long and short positions that’s a great way to easily remember it. Just ignore PV(X) for synthetics as you ignore the fact you need cash when the options expire.

This is what I normally do as well - I physically write down c + x = p + s, and proceed to physically crossing out PV(X). The rest is just simple algebra (in our case here, we want -s). c - p = s, -s = p - c.