Dependency of an option price on time till expiry
$begingroup$
I am trying to seek satisfaction when it comes to understanding why the price of an option is dependent on the time until expiry.
I have read that the longer till expiration, the more time available for movement in the underlying stock price, and more chance of being on the desirable side of the strike price (assuming positive drift), so this should mean that the option price is higher than when we have shorter times till expiration.
But if the drift was a larger positive, this should mean that the price of the option is greater as we have even more chance now of landing in the money. But this isn't true since the drift does not affect the option price, telling me that the reasoning above isn't quite right.
Could someone please shed some light on how to understand how the time till expiration affects the option price.
options option-pricing european-options
edited Jan 28 at 20:54
LocalVolatility
5,49631328
asked Jan 28 at 17:53
DLBDLB
503
$endgroup$
add a comment |
$begingroup$
I am trying to seek satisfaction when it comes to understanding why the price of an option is dependent on the time until expiry.
I have read that the longer till expiration, the more time available for movement in the underlying stock price, and more chance of being on the desirable side of the strike price (assuming positive drift), so this should mean that the option price is higher than when we have shorter times till expiration.
But if the drift was a larger positive, this should mean that the price of the option is greater as we have even more chance now of landing in the money. But this isn't true since the drift does not affect the option price, telling me that the reasoning above isn't quite right.
Could someone please shed some light on how to understand how the time till expiration affects the option price.
options option-pricing european-options
edited Jan 28 at 20:54
LocalVolatility
5,49631328
asked Jan 28 at 17:53
DLBDLB
503
$endgroup$
2
$begingroup$
The more time until expiration the more "spread out" will be the stock price at maturity. Tomorrow the price of AAPL wil be a couple of percent above/below today, but a year from now it could be dozens of percent above below. To a stockholder this does not matter much because the possibility of a rise is matched by the possibility of a fall. But the payoff to a call is asymmetric, you profit greatly from a big rise but you do not suffer much from a big fall since the payoff is zero. Therefore the call holder benefits from a wider dispersion of stock prices at maturity.
$endgroup$
– Alex C
Jan 28 at 18:05
add a comment |
$begingroup$
I am trying to seek satisfaction when it comes to understanding why the price of an option is dependent on the time until expiry.
I have read that the longer till expiration, the more time available for movement in the underlying stock price, and more chance of being on the desirable side of the strike price (assuming positive drift), so this should mean that the option price is higher than when we have shorter times till expiration.
But if the drift was a larger positive, this should mean that the price of the option is greater as we have even more chance now of landing in the money. But this isn't true since the drift does not affect the option price, telling me that the reasoning above isn't quite right.
Could someone please shed some light on how to understand how the time till expiration affects the option price.
options option-pricing european-options
edited Jan 28 at 20:54
LocalVolatility
5,49631328
asked Jan 28 at 17:53
DLBDLB
503
$endgroup$
I am trying to seek satisfaction when it comes to understanding why the price of an option is dependent on the time until expiry.
I have read that the longer till expiration, the more time available for movement in the underlying stock price, and more chance of being on the desirable side of the strike price (assuming positive drift), so this should mean that the option price is higher than when we have shorter times till expiration.
But if the drift was a larger positive, this should mean that the price of the option is greater as we have even more chance now of landing in the money. But this isn't true since the drift does not affect the option price, telling me that the reasoning above isn't quite right.
Could someone please shed some light on how to understand how the time till expiration affects the option price.
options option-pricing european-options
options option-pricing european-options
edited Jan 28 at 20:54
LocalVolatility
5,49631328
asked Jan 28 at 17:53
DLBDLB
503
edited Jan 28 at 20:54
LocalVolatility
5,49631328
asked Jan 28 at 17:53
DLBDLB
503
edited Jan 28 at 20:54
LocalVolatility
5,49631328
edited Jan 28 at 20:54
LocalVolatility
5,49631328
edited Jan 28 at 20:54
LocalVolatility
5,49631328
5,49631328
asked Jan 28 at 17:53
DLBDLB
503
asked Jan 28 at 17:53
DLBDLB
503
asked Jan 28 at 17:53
DLBDLB
503
503
2
$begingroup$
The more time until expiration the more "spread out" will be the stock price at maturity. Tomorrow the price of AAPL wil be a couple of percent above/below today, but a year from now it could be dozens of percent above below. To a stockholder this does not matter much because the possibility of a rise is matched by the possibility of a fall. But the payoff to a call is asymmetric, you profit greatly from a big rise but you do not suffer much from a big fall since the payoff is zero. Therefore the call holder benefits from a wider dispersion of stock prices at maturity.
$endgroup$
– Alex C
Jan 28 at 18:05
add a comment |
2
$begingroup$
The more time until expiration the more "spread out" will be the stock price at maturity. Tomorrow the price of AAPL wil be a couple of percent above/below today, but a year from now it could be dozens of percent above below. To a stockholder this does not matter much because the possibility of a rise is matched by the possibility of a fall. But the payoff to a call is asymmetric, you profit greatly from a big rise but you do not suffer much from a big fall since the payoff is zero. Therefore the call holder benefits from a wider dispersion of stock prices at maturity.
$endgroup$
– Alex C
Jan 28 at 18:05
2
2
$begingroup$
The more time until expiration the more "spread out" will be the stock price at maturity. Tomorrow the price of AAPL wil be a couple of percent above/below today, but a year from now it could be dozens of percent above below. To a stockholder this does not matter much because the possibility of a rise is matched by the possibility of a fall. But the payoff to a call is asymmetric, you profit greatly from a big rise but you do not suffer much from a big fall since the payoff is zero. Therefore the call holder benefits from a wider dispersion of stock prices at maturity.
$endgroup$
– Alex C
Jan 28 at 18:05
$begingroup$
The more time until expiration the more "spread out" will be the stock price at maturity. Tomorrow the price of AAPL wil be a couple of percent above/below today, but a year from now it could be dozens of percent above below. To a stockholder this does not matter much because the possibility of a rise is matched by the possibility of a fall. But the payoff to a call is asymmetric, you profit greatly from a big rise but you do not suffer much from a big fall since the payoff is zero. Therefore the call holder benefits from a wider dispersion of stock prices at maturity.
$endgroup$
– Alex C
Jan 28 at 18:05
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You've tagged this with 'black-scholes' but you don't have to make the assumptions of the Black-Scholes-Merton model to understand why the option price with time to expiry.
Consider this example: Consider 2 ATM put options on a stock with a time to expiry of one month and one year with some strike price. The maximum pay-out is achieved when the company goes bust, what is more likely, and which option is thus more valuable, that the company goes bust within a month or within a year?
More generally, the price of an ATM or OTM option increases as uncertainty about the price at expiry increases.
More uncertainty allows the stock price to move further away from the strike in the profitable correction which is increases the value, for a call the value can grow without bound. It also can move the other way in which case the option becomes less valuable but never zero or less.
For ITM put options, time can work against the option holder. The stock has a hard lower bound so the upside for the option holder is limited but it could make a recovery in which case the option decreases in value.
answered Jan 28 at 18:08
Bob Jansen♦Bob Jansen
3,51252245
$endgroup$
$begingroup$
Regarding the example you provided, I have seen a lot of graphed solutions to the Black Scholes PDE and they indicate a shorter time till expiration comes with a greater price?
$endgroup$
– DLB
Jan 28 at 18:32
$begingroup$
Could you show such a graph, all else equal price should increase with time till expiration.
$endgroup$
– Bob Jansen♦
Jan 28 at 18:35
$begingroup$
Figure 2 in the link math.unl.edu/~sdunbar1/MathematicalFinance/Lessons/BlackScholes/… shows the value of a put option at different times.
$endgroup$
– DLB
Jan 28 at 18:39
$begingroup$
I hadn't considered that case, updated the answer.
$endgroup$
– Bob Jansen♦
Jan 28 at 18:54
$begingroup$
Is there a reason we don't see this same case structure when we consider call options?
$endgroup$
– DLB
Jan 28 at 19:04
|
show 1 more comment
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "204"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fquant.stackexchange.com%2fquestions%2f43748%2fdependency-of-an-option-price-on-time-till-expiry%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You've tagged this with 'black-scholes' but you don't have to make the assumptions of the Black-Scholes-Merton model to understand why the option price with time to expiry.
Consider this example: Consider 2 ATM put options on a stock with a time to expiry of one month and one year with some strike price. The maximum pay-out is achieved when the company goes bust, what is more likely, and which option is thus more valuable, that the company goes bust within a month or within a year?
More generally, the price of an ATM or OTM option increases as uncertainty about the price at expiry increases.
More uncertainty allows the stock price to move further away from the strike in the profitable correction which is increases the value, for a call the value can grow without bound. It also can move the other way in which case the option becomes less valuable but never zero or less.
For ITM put options, time can work against the option holder. The stock has a hard lower bound so the upside for the option holder is limited but it could make a recovery in which case the option decreases in value.
answered Jan 28 at 18:08
Bob Jansen♦Bob Jansen
3,51252245
$endgroup$
$begingroup$
Regarding the example you provided, I have seen a lot of graphed solutions to the Black Scholes PDE and they indicate a shorter time till expiration comes with a greater price?
$endgroup$
– DLB
Jan 28 at 18:32
$begingroup$
Could you show such a graph, all else equal price should increase with time till expiration.
$endgroup$
– Bob Jansen♦
Jan 28 at 18:35
$begingroup$
Figure 2 in the link math.unl.edu/~sdunbar1/MathematicalFinance/Lessons/BlackScholes/… shows the value of a put option at different times.
$endgroup$
– DLB
Jan 28 at 18:39
$begingroup$
I hadn't considered that case, updated the answer.
$endgroup$
– Bob Jansen♦
Jan 28 at 18:54
$begingroup$
Is there a reason we don't see this same case structure when we consider call options?
$endgroup$
– DLB
Jan 28 at 19:04
|
show 1 more comment
$begingroup$
You've tagged this with 'black-scholes' but you don't have to make the assumptions of the Black-Scholes-Merton model to understand why the option price with time to expiry.
Consider this example: Consider 2 ATM put options on a stock with a time to expiry of one month and one year with some strike price. The maximum pay-out is achieved when the company goes bust, what is more likely, and which option is thus more valuable, that the company goes bust within a month or within a year?
More generally, the price of an ATM or OTM option increases as uncertainty about the price at expiry increases.
More uncertainty allows the stock price to move further away from the strike in the profitable correction which is increases the value, for a call the value can grow without bound. It also can move the other way in which case the option becomes less valuable but never zero or less.
For ITM put options, time can work against the option holder. The stock has a hard lower bound so the upside for the option holder is limited but it could make a recovery in which case the option decreases in value.
answered Jan 28 at 18:08
Bob Jansen♦Bob Jansen
3,51252245
$endgroup$
$begingroup$
Regarding the example you provided, I have seen a lot of graphed solutions to the Black Scholes PDE and they indicate a shorter time till expiration comes with a greater price?
$endgroup$
– DLB
Jan 28 at 18:32
$begingroup$
Could you show such a graph, all else equal price should increase with time till expiration.
$endgroup$
– Bob Jansen♦
Jan 28 at 18:35
$begingroup$
Figure 2 in the link math.unl.edu/~sdunbar1/MathematicalFinance/Lessons/BlackScholes/… shows the value of a put option at different times.
$endgroup$
– DLB
Jan 28 at 18:39
$begingroup$
I hadn't considered that case, updated the answer.
$endgroup$
– Bob Jansen♦
Jan 28 at 18:54
$begingroup$
Is there a reason we don't see this same case structure when we consider call options?
$endgroup$
– DLB
Jan 28 at 19:04
|
show 1 more comment
$begingroup$
You've tagged this with 'black-scholes' but you don't have to make the assumptions of the Black-Scholes-Merton model to understand why the option price with time to expiry.
Consider this example: Consider 2 ATM put options on a stock with a time to expiry of one month and one year with some strike price. The maximum pay-out is achieved when the company goes bust, what is more likely, and which option is thus more valuable, that the company goes bust within a month or within a year?
More generally, the price of an ATM or OTM option increases as uncertainty about the price at expiry increases.
More uncertainty allows the stock price to move further away from the strike in the profitable correction which is increases the value, for a call the value can grow without bound. It also can move the other way in which case the option becomes less valuable but never zero or less.
For ITM put options, time can work against the option holder. The stock has a hard lower bound so the upside for the option holder is limited but it could make a recovery in which case the option decreases in value.
answered Jan 28 at 18:08
Bob Jansen♦Bob Jansen
3,51252245
$endgroup$
You've tagged this with 'black-scholes' but you don't have to make the assumptions of the Black-Scholes-Merton model to understand why the option price with time to expiry.
Consider this example: Consider 2 ATM put options on a stock with a time to expiry of one month and one year with some strike price. The maximum pay-out is achieved when the company goes bust, what is more likely, and which option is thus more valuable, that the company goes bust within a month or within a year?
More generally, the price of an ATM or OTM option increases as uncertainty about the price at expiry increases.
More uncertainty allows the stock price to move further away from the strike in the profitable correction which is increases the value, for a call the value can grow without bound. It also can move the other way in which case the option becomes less valuable but never zero or less.
For ITM put options, time can work against the option holder. The stock has a hard lower bound so the upside for the option holder is limited but it could make a recovery in which case the option decreases in value.
answered Jan 28 at 18:08
Bob Jansen♦Bob Jansen
3,51252245
edited Jan 28 at 18:54
answered Jan 28 at 18:08
Bob Jansen♦Bob Jansen
3,51252245
answered Jan 28 at 18:08
Bob Jansen♦Bob Jansen
3,51252245
answered Jan 28 at 18:08
Bob Jansen♦Bob Jansen
3,51252245
3,51252245
$begingroup$
Regarding the example you provided, I have seen a lot of graphed solutions to the Black Scholes PDE and they indicate a shorter time till expiration comes with a greater price?
$endgroup$
– DLB
Jan 28 at 18:32
$begingroup$
Could you show such a graph, all else equal price should increase with time till expiration.
$endgroup$
– Bob Jansen♦
Jan 28 at 18:35
$begingroup$
Figure 2 in the link math.unl.edu/~sdunbar1/MathematicalFinance/Lessons/BlackScholes/… shows the value of a put option at different times.
$endgroup$
– DLB
Jan 28 at 18:39
$begingroup$
I hadn't considered that case, updated the answer.
$endgroup$
– Bob Jansen♦
Jan 28 at 18:54
$begingroup$
Is there a reason we don't see this same case structure when we consider call options?
$endgroup$
– DLB
Jan 28 at 19:04
|
show 1 more comment
$begingroup$
Regarding the example you provided, I have seen a lot of graphed solutions to the Black Scholes PDE and they indicate a shorter time till expiration comes with a greater price?
$endgroup$
– DLB
Jan 28 at 18:32
$begingroup$
Could you show such a graph, all else equal price should increase with time till expiration.
$endgroup$
– Bob Jansen♦
Jan 28 at 18:35
$begingroup$
Figure 2 in the link math.unl.edu/~sdunbar1/MathematicalFinance/Lessons/BlackScholes/… shows the value of a put option at different times.
$endgroup$
– DLB
Jan 28 at 18:39
$begingroup$
I hadn't considered that case, updated the answer.
$endgroup$
– Bob Jansen♦
Jan 28 at 18:54
$begingroup$
Is there a reason we don't see this same case structure when we consider call options?
$endgroup$
– DLB
Jan 28 at 19:04
$begingroup$
Regarding the example you provided, I have seen a lot of graphed solutions to the Black Scholes PDE and they indicate a shorter time till expiration comes with a greater price?
$endgroup$
– DLB
Jan 28 at 18:32
$begingroup$
Regarding the example you provided, I have seen a lot of graphed solutions to the Black Scholes PDE and they indicate a shorter time till expiration comes with a greater price?
$endgroup$
– DLB
Jan 28 at 18:32
$begingroup$
Could you show such a graph, all else equal price should increase with time till expiration.
$endgroup$
– Bob Jansen♦
Jan 28 at 18:35
$begingroup$
Could you show such a graph, all else equal price should increase with time till expiration.
$endgroup$
– Bob Jansen♦
Jan 28 at 18:35
$begingroup$
Figure 2 in the link math.unl.edu/~sdunbar1/MathematicalFinance/Lessons/BlackScholes/… shows the value of a put option at different times.
$endgroup$
– DLB
Jan 28 at 18:39
$begingroup$
Figure 2 in the link math.unl.edu/~sdunbar1/MathematicalFinance/Lessons/BlackScholes/… shows the value of a put option at different times.
$endgroup$
– DLB
Jan 28 at 18:39
$begingroup$
I hadn't considered that case, updated the answer.
$endgroup$
– Bob Jansen♦
Jan 28 at 18:54
$begingroup$
I hadn't considered that case, updated the answer.
$endgroup$
– Bob Jansen♦
Jan 28 at 18:54
$begingroup$
Is there a reason we don't see this same case structure when we consider call options?
$endgroup$
– DLB
Jan 28 at 19:04
$begingroup$
Is there a reason we don't see this same case structure when we consider call options?
$endgroup$
– DLB
Jan 28 at 19:04
|
show 1 more comment
Thanks for contributing an answer to Quantitative Finance Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fquant.stackexchange.com%2fquestions%2f43748%2fdependency-of-an-option-price-on-time-till-expiry%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
The more time until expiration the more "spread out" will be the stock price at maturity. Tomorrow the price of AAPL wil be a couple of percent above/below today, but a year from now it could be dozens of percent above below. To a stockholder this does not matter much because the possibility of a rise is matched by the possibility of a fall. But the payoff to a call is asymmetric, you profit greatly from a big rise but you do not suffer much from a big fall since the payoff is zero. Therefore the call holder benefits from a wider dispersion of stock prices at maturity.
$endgroup$
– Alex C
Jan 28 at 18:05