Can increase in volatility reduce the price of a deeply in-the-money European put?Is the price of European put option monotone in volatility if we replace BM in Black-Scholes with a general Levy process?Approximation of an option priceCox-Ross-Rubinstein - getting volatilityParadox in option expiry as volatility goes to infinityQuoting options with reference price and deltaEquity Options - “How do I build a forward simulation model with regards to shocks in spot pricing and IV?”Perpetual Put vs European PutWhat are the main problems for calculating the implied volatility of in the money American put options?Computing option price with rates onlyWhat is the cause of a “broken” volatility surface?
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Can increase in volatility reduce the price of a deeply in-the-money European put?
Is the price of European put option monotone in volatility if we replace BM in Black-Scholes with a general Levy process?Approximation of an option priceCox-Ross-Rubinstein - getting volatilityParadox in option expiry as volatility goes to infinityQuoting options with reference price and deltaEquity Options - “How do I build a forward simulation model with regards to shocks in spot pricing and IV?”Perpetual Put vs European PutWhat are the main problems for calculating the implied volatility of in the money American put options?Computing option price with rates onlyWhat is the cause of a “broken” volatility surface?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
Hull states that option prices increase with an increase in volatility.
I think that statement could be false in a specific scenario: when we are considering a deeply in-the-money European put option.
Since we are deeply in-the-money, the price of the underlying would be close to zero. Since the price of the underlying can't be negative, the effect of volatility would be asymetric: it would be more likely for the share price to recover than to fall anymore, simply because there isn't a lot of scope for a fall to happen from an already near-zero share price.
So a higher volatility is more likely to lead to a recovery of the share price, reducing the payoff, in turn reducing the price of the European put.
Is my reasoning wrong? Thanks in advance!
volatility european-options put
asked 9 hours ago
Dhruv GuptaDhruv Gupta
795 bronze badges
$endgroup$
add a comment |
$begingroup$
Hull states that option prices increase with an increase in volatility.
I think that statement could be false in a specific scenario: when we are considering a deeply in-the-money European put option.
Since we are deeply in-the-money, the price of the underlying would be close to zero. Since the price of the underlying can't be negative, the effect of volatility would be asymetric: it would be more likely for the share price to recover than to fall anymore, simply because there isn't a lot of scope for a fall to happen from an already near-zero share price.
So a higher volatility is more likely to lead to a recovery of the share price, reducing the payoff, in turn reducing the price of the European put.
Is my reasoning wrong? Thanks in advance!
volatility european-options put
asked 9 hours ago
Dhruv GuptaDhruv Gupta
795 bronze badges
$endgroup$
add a comment |
$begingroup$
Hull states that option prices increase with an increase in volatility.
I think that statement could be false in a specific scenario: when we are considering a deeply in-the-money European put option.
Since we are deeply in-the-money, the price of the underlying would be close to zero. Since the price of the underlying can't be negative, the effect of volatility would be asymetric: it would be more likely for the share price to recover than to fall anymore, simply because there isn't a lot of scope for a fall to happen from an already near-zero share price.
So a higher volatility is more likely to lead to a recovery of the share price, reducing the payoff, in turn reducing the price of the European put.
Is my reasoning wrong? Thanks in advance!
volatility european-options put
asked 9 hours ago
Dhruv GuptaDhruv Gupta
795 bronze badges
$endgroup$
Hull states that option prices increase with an increase in volatility.
I think that statement could be false in a specific scenario: when we are considering a deeply in-the-money European put option.
Since we are deeply in-the-money, the price of the underlying would be close to zero. Since the price of the underlying can't be negative, the effect of volatility would be asymetric: it would be more likely for the share price to recover than to fall anymore, simply because there isn't a lot of scope for a fall to happen from an already near-zero share price.
So a higher volatility is more likely to lead to a recovery of the share price, reducing the payoff, in turn reducing the price of the European put.
Is my reasoning wrong? Thanks in advance!
volatility european-options put
volatility european-options put
asked 9 hours ago
Dhruv GuptaDhruv Gupta
795 bronze badges
asked 9 hours ago
Dhruv GuptaDhruv Gupta
795 bronze badges
asked 9 hours ago
Dhruv GuptaDhruv Gupta
795 bronze badges
asked 9 hours ago
Dhruv GuptaDhruv Gupta
795 bronze badges
asked 9 hours ago
Dhruv GuptaDhruv Gupta
795 bronze badges
795 bronze badges
add a comment |
add a comment |
1 Answer
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$begingroup$
If you hold an option, you're always vega long, i.e. if volatility increases, your position increases as well - regardless of moneyness and the option type (put or call). Note firstly that by the model-free put-call parity, put and call options have the same vega (i.e. changes in volatility affect put and call prices in an identical way).
Let now $Kgg S_t$, then your put option is deep ITM but a corresponding call option would be deep OTM and what about the logic ''the call has nothing to lose and can only win, so increasing volatility should increase the call price'' but that would then also imply increasing put prices.
In the Black-Scholes model,
beginalign*
mathrmVega &= S_te^-qTvarphi(d_1)sqrtT-t \
&= Ke^-r(T-t)varphi(d_2)sqrtT-t
endalign*
which is always positive. Here, $varphi$ denotes the probability distribution function of a normally distributed random variable.
answered 6 hours ago
KeSchnKeSchn
91810 bronze badges
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
If you hold an option, you're always vega long, i.e. if volatility increases, your position increases as well - regardless of moneyness and the option type (put or call). Note firstly that by the model-free put-call parity, put and call options have the same vega (i.e. changes in volatility affect put and call prices in an identical way).
Let now $Kgg S_t$, then your put option is deep ITM but a corresponding call option would be deep OTM and what about the logic ''the call has nothing to lose and can only win, so increasing volatility should increase the call price'' but that would then also imply increasing put prices.
In the Black-Scholes model,
beginalign*
mathrmVega &= S_te^-qTvarphi(d_1)sqrtT-t \
&= Ke^-r(T-t)varphi(d_2)sqrtT-t
endalign*
which is always positive. Here, $varphi$ denotes the probability distribution function of a normally distributed random variable.
answered 6 hours ago
KeSchnKeSchn
91810 bronze badges
$endgroup$
add a comment |
$begingroup$
If you hold an option, you're always vega long, i.e. if volatility increases, your position increases as well - regardless of moneyness and the option type (put or call). Note firstly that by the model-free put-call parity, put and call options have the same vega (i.e. changes in volatility affect put and call prices in an identical way).
Let now $Kgg S_t$, then your put option is deep ITM but a corresponding call option would be deep OTM and what about the logic ''the call has nothing to lose and can only win, so increasing volatility should increase the call price'' but that would then also imply increasing put prices.
In the Black-Scholes model,
beginalign*
mathrmVega &= S_te^-qTvarphi(d_1)sqrtT-t \
&= Ke^-r(T-t)varphi(d_2)sqrtT-t
endalign*
which is always positive. Here, $varphi$ denotes the probability distribution function of a normally distributed random variable.
answered 6 hours ago
KeSchnKeSchn
91810 bronze badges
$endgroup$
add a comment |
$begingroup$
If you hold an option, you're always vega long, i.e. if volatility increases, your position increases as well - regardless of moneyness and the option type (put or call). Note firstly that by the model-free put-call parity, put and call options have the same vega (i.e. changes in volatility affect put and call prices in an identical way).
Let now $Kgg S_t$, then your put option is deep ITM but a corresponding call option would be deep OTM and what about the logic ''the call has nothing to lose and can only win, so increasing volatility should increase the call price'' but that would then also imply increasing put prices.
In the Black-Scholes model,
beginalign*
mathrmVega &= S_te^-qTvarphi(d_1)sqrtT-t \
&= Ke^-r(T-t)varphi(d_2)sqrtT-t
endalign*
which is always positive. Here, $varphi$ denotes the probability distribution function of a normally distributed random variable.
answered 6 hours ago
KeSchnKeSchn
91810 bronze badges
$endgroup$
If you hold an option, you're always vega long, i.e. if volatility increases, your position increases as well - regardless of moneyness and the option type (put or call). Note firstly that by the model-free put-call parity, put and call options have the same vega (i.e. changes in volatility affect put and call prices in an identical way).
Let now $Kgg S_t$, then your put option is deep ITM but a corresponding call option would be deep OTM and what about the logic ''the call has nothing to lose and can only win, so increasing volatility should increase the call price'' but that would then also imply increasing put prices.
In the Black-Scholes model,
beginalign*
mathrmVega &= S_te^-qTvarphi(d_1)sqrtT-t \
&= Ke^-r(T-t)varphi(d_2)sqrtT-t
endalign*
which is always positive. Here, $varphi$ denotes the probability distribution function of a normally distributed random variable.
answered 6 hours ago
KeSchnKeSchn
91810 bronze badges
edited 6 hours ago
answered 6 hours ago
KeSchnKeSchn
91810 bronze badges
answered 6 hours ago
KeSchnKeSchn
91810 bronze badges
answered 6 hours ago
KeSchnKeSchn
91810 bronze badges
91810 bronze badges
add a comment |
add a comment |
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