7
$\begingroup$

I have a negligible amount of money (\$5000) that I would like to invest in a stock. I would like to buy the stock at some point in the next year, and get the lowest possible price.

I would like to accomplish this by placing a limit order at \$X to buy (5000/$X) shares of stock A, today, with, say, a 98% chance that the price dips below $X and the order executes at some point in the next year. I don't need the calculation to be perfect, but I'm new to financial modeling and don't really know where to start looking. I started looking at the prices of out-of-the-money puts, though these are based on the profit you can make from selling the stock at that value, they don't really express the likelihood of the price dipping below that amount.

What's a simple model, or existing market/contract somewhere, that I can use to model the probability of a stock dipping or jumping by some amount over a given time period?

$\endgroup$
  • 3
    $\begingroup$ My suggestion would be: assume a rate of return $\mu$ and a volatility $\sigma$. Run a Montecarlo simulation of GBM for 1 year to find out at what percentage of the initial stock price $S_0$ the order would have to be to be hit X percent of the time. Make a table showing these for various levels of X. $\endgroup$ – Alex C Mar 19 '16 at 19:44
  • $\begingroup$ My suggestion is the market is random and nothing can be predicted with any degree of probability. It sounds like you should try reading reminiscences of a stock operator $\endgroup$ – rupweb Mar 21 '16 at 9:39
  • 1
    $\begingroup$ @rupweb In the extreme philosophical sense, probability doesn't exist at all: an event either occurs or fails to occur, so your comment applies to pretty much any question on any of the sites that deal with non-theoretical probability. In most such cases, however, we're talking about a theoretical probability based on assumptions we postulate to be true. $\endgroup$ – barrycarter Mar 21 '16 at 14:14
  • $\begingroup$ @barrycarter hahaha, well, people think they can predict asset prices... here we want to know some probability that a stock A reaches a price $X. I don't think that is predictable. I think a better question is whether we can find a probability that all stocks A* will reach all equivalent prices $X*... $\endgroup$ – rupweb Mar 21 '16 at 21:35
8
+100
$\begingroup$

My Answer

You should set your limit order to: $s (v+1)^{-0.0314192 \sqrt{t}}$ where $s$ is the current price, $t$ is the time in years you're willing to wait, and $v$ is the annual volatility as a percentage.

If you want to be $p$ percent sure (instead of 0.98), set your limit order to:

$s (v+1)^{-\sqrt{\pi } \sqrt{t} \text{erf}^{-1}(1-p)}$

Of course, this is based on many assumptions and disclaimers later in this message.

Other Answers

It turns out this question has been studied extensively, and there are some papers on it:

http://fiquant.mas.ecp.fr/wp-content/uploads/2015/10/Limit-Order-Book-modelling.pdf

http://arxiv.org/pdf/1311.5661

I'll use a much simpler model (see disclaimers at end of message).

Example

If a stock has a volatility of 15%, that means there's a 68% chance it's price after 1 year will be between 87% and 115% of its current price. Note that the lower limit is 87% (= 1/1.15), not 85%.

Overall, the price probability for a stock with volatility 15% forms this bell curve:

enter image description here

Note that:

  • Because volatility is inherently based on logrithms, the tick marks aren't evenly numbered, and aren't symmetric. For example, the numbers +65% and -39% are symmetric because it takes a 39% loss to offset a 65% gain and vice versa. In other words: (1+ (-39/100))*(1+ (65/100)) is approximately one.

  • The parenthesized numbers under the x axis (for this and the following graphs) refer to change in the logarithm of the security's price. These are evenly numbered and we will use them in the "General Case" section.

  • The labels on the y axis are relative to each other and don't refer to percentages.

Of course, this isn't the probability curve you're looking for: I drew it just for reference.

Instead, let's look at the probability distribution of the minimum value over the next year for our 15% volatility stock.

enter image description here

The same caveats apply to this graph as the previous one.

Suppose you set your limit order at 5% below the current price (ie, 95% of its current price). There is a ~77% chance your order will be filled:

enter image description here

You can also see this using the cumulative distribution function (CDF):

enter image description here

In this case, the y values do represent percentages, namely the cumulative percentage change that the stock's lowest value will the percentage value on the x axis.

For this volatility, if you want be 98% sure you order is filled, you could only set your limit order to 0.44% below the current price.

General Case

Of course, that was for a specific volatility over a specific period of time.

In general, a volatility of v% means the stock is ~68% (1 standard deviation) likely to remain within v% of its current price in the next year. More conveniently, it means the logarithm of the price is 68% likely to remain within (plus/minus) $\log (v+1)$ of its current value (within the next year). For example, a volatility of 15% means the log of the stock price is 68% likely to remain within .1398 of its current value, since $e^{0.1398}$ is approximately $1.15$

More generally, the $\log (\text{price})$ one year from now has a normal distribution with mean $\log (\text{price})$ and standard deviation $\log (v+1)$.

Thus, the change in the $\log (\text{price})$ for one year is normally distributed with a mean of 0 and a standard deviation of $\log (v+1)$.

A standard deviation of $\log (v+1)$ translates to a variance of $\log ^2(v+1)$. Since the variance of a process like this scales linearlly, the variance for $t$ years is given by $t \log ^2(v+1)$ and the standard deviation for $t$ years is given by $\sqrt{t} \log (v+1)$.

Thus, the change in $\log (\text{price})$ for time $t$ has a normal distribution with mean 0 and standard deviation $\sqrt{t} \log (v+1)$.

As noted below in another section, this means the minimum (most negative) value of this change has a halfnormal distribution with parameter $\frac{1}{\sqrt{t} \log (v+1)}$

The cumulative distribution of a halfnormal distribution with parameter $\frac{1}{\sqrt{t} \log (v+1)}$ evaluted at x>0 (the only place the halfnormal distribution is non-zero) is:

$\text{erf}\left(\frac{x}{\sqrt{\pi } \sqrt{t} \log (v+1)}\right)$

where erf() is the standard error function.

Note that when we draw this cumulative distribution for volatility 15% above, letting the x axis be "change in $\log (\text{price})$ (instead of the percentage change in price), the x axis looks more like we expect.

If our limit order is $\lambda$% of the current price (meaning it's $\lambda s$ where $s$ is the current price), it will only be hit if the $\log (\text{price})$ moves more than $\left| \log (\lambda ) \right|$ (note that we need the absolute values since we're measuring the absolute change in $\log (\text{price})$, which is always positive). The chance of that happening is:

$ 1-\text{erf}\left(\frac{\left| \log (\lambda ) \right|}{\sqrt{\pi } \sqrt{t} \log (v+1)}\right) $

Note that we need the "1-" since we're looking for the probability the $\log (\text{price})$ moves more than the given amount.

Of course, in this case, we're given the probability and asked to solve for the limit price. Using $p$ as the probability we find:

$\lambda \to (v+1)^{-\sqrt{\pi } \sqrt{t} \text{erf}^{-1}(1-p)}$

and the price is thus:

$s (v+1)^{-\sqrt{\pi } \sqrt{t} \text{erf}^{-1}(1-p)}$

as in the answer section. Substituting 0.98 for p, we have:

$s (v+1)^{-0.0314192 \sqrt{t}}$

as noted for this specific example.

Research and "derivation"

It turns out this is a well-known problem and has been studied extensively:

https://stackexchange.com/search?q=brownian+halfnormal

  • It can also be regarded as the running maximum value of a random walk:

https://stackexchange.com/search?q=brownian+halfnormal

What is the fair price of this option?

Probability of touching

If you use Mathematica (or just want to read even more about this subject), you might look at my:

the latter of which computes the probability that a stock price will be between two given values at two given times (ie, the fair value of an O&A "box option") but can be used to answer your question in the limiting case. See also: https://money.stackexchange.com/questions/4312

Disclaimers and Notices

I made several simplifying assumptions above:

  • As noted in the references given in "Other Answers" above, the more a stock's price decreases, the less likely it is to decrease further. Why? Other people place limit orders, and the further down the stock gets from its starting price, the more limit orders will be triggered. Generally, the volume of limit orders also increases as the stock price goes down. In other words, the limit orders act as a "buffer", slowing the rate at which a stock's price drops. The simple model I use does not account for this.

  • Conversely, I also ignore the "volatility smile", which suggests the exact opposite: that a larger change in price is more likely than what the normal distribution would yield, which means that extreme prices are more likely that those given by the halfnormal distribution.

  • The two points above aren't necessarily contradictory: under normal conditions, the "limit order book" buffers price changes, but during unusual circumstances (such as major news), the price can change dramatically.

  • I also assume that once a stock reaches your limit price, your order will be triggered. However, if there are several orders at that price, the larger orders will trigger first, and the stock price may rise again before your limit order is triggered at all.

  • Since this is a limit order and not an option, the risk-free interest rate is not an issue: I assume you earn the risk-free interest rate until the order is filled.

  • If you don't earn the risk-free interest rate while waiting, note that the small gain you get from the limit order may be offset by the loss of interest.

Items Not Appearing in This Answer

Although the inverse error function is well known and "easy" to compute, I was going to include an approximation, but felt that might exceed the scope of the question.

$\endgroup$
  • $\begingroup$ nice answer. all about the volatility of a stock. presumably the volatility of the whole market is priced in, hmm... $\endgroup$ – rupweb Mar 29 '16 at 15:24
  • $\begingroup$ @rupweb Only in the sense that the volatility of a given stock may change with market volatility, but I don't make any specific reference to market volatility. $\endgroup$ – barrycarter Mar 29 '16 at 18:01
  • $\begingroup$ thanks for this! is erf^-1 the same as erfc, e.g. here? golang.org/pkg/math/#Erfc $\endgroup$ – Kevin Burke Apr 4 '16 at 0:39
  • $\begingroup$ Erfc s 1-erf. erf^-1 means the inverse of the erf function. mathworld.wolfram.com/InverseErf.html $\endgroup$ – barrycarter Apr 4 '16 at 1:26
  • $\begingroup$ do you have an idea about we can add to this formula a trend param ? so we can bias the result? $\endgroup$ – mahieddine May 18 '18 at 23:26
4
$\begingroup$

One way to do this is a simple Monte-Carlo simulation. There are formulae you can use to get the likelihood of a stock being below a price if you know the stock's volatility and time frame - see for example this question. For an unknown time frame, the Monte-Carlo method is (IMO) simpler than the mathematics.

You would simply run a number of simulations using your expectation of volatility, or the market's (e.g. 1 year put option's implied vol at your given strike) of the stock price's movement over one year. Then the percentage chance of it moving below a price over a year, is simply the number of series that move below that price (at any point) during the year / total number of sims:

$P(S\ moves\ below\ K)= \frac{Number\ that\ move \ below\ K}{Total\ Number\ of\ Sims}$

This will converge to the actual percentage for standard Geometric Brownian Motion or whatever your Monte Carlo does, on the assumption that the stock follows the random walk hypothesis. You can of course tweak it for your beliefs of stock drift, and/or future volatility.

Note: If you need help with the Monte-Carlo sim, I can probably post or direct you to code in whatever is your preferred language.

$\endgroup$
2
$\begingroup$

This paper is a microstructure paper (http://papers.ssrn.com/sol3/papers.cfm?abstract_id=2185452) that you may be looking at.

We will assume the complex dynamics originated by the interaction between sell- and buy- side (limit and market orders) can be modelled in terms of a continuous stochastic process Mt, the microprice, which we assume to be a Brownian motion,

$$d M_t = \sigma_t dW_t$$

The microprice Mt represents the fair value of the security. Since the security can only trade at the discrete prices ${S_i}$, the microprice itself is not directly observable in the market place. Nonetheless, a commonly used empirical proxy for the microprice as a particular weighted average of the quote prices is inspired by the following argument. Denote by $S_t^+$ and $S_t^−$ the best bid and offer prices at time t, $V_t^+$ and $V_t^−$ the corresponding quantities posted in the order book. Clearly $S_t^+ \leq M_t \leq S_t^−$. As the microprice approaches one side of the market, let’s say St+, more buy-side investors will be willing to pay the reduced price of immediacy, the spread between the bid and the fair value $M_t −S_t^+$, in order to sell the desired amount immediately. In turn, this will result in additional sell market orders hitting the bid. At the same time it becomes more profitable for dealers to sell at the best offer $S_t^−$, so the corresponding offered size $V_t$− will increase. The reverse is true as the microprice goes up: now it looks more attractive for investor to buy at the offered price $S_t^−$ as the cost of immediate execution $S_t^− −M_t$ is reduced, so we can expect additional buy market orders lifting the offer, and more dealers will try to buy at the best bid $S_t^+$ thereby increasing $V_t^+$. Eventually, as the microprice touches a tradeable price, let’s say the bid $M_t = S_t^+$, the demand from investor to sell at prices above $S_t^+$ is exhausted and no additional sell market orders are expected. At the same time, dealers will not be willing to buy any additional units at $S_t^+$ because they will not get any compensation for the trade. Therefore $V_t^+$ goes to zero as Mt approaches $S_t^+$ from above and similarly $V_t^−$ goes to zero as $M_t$ approaches $S_t^−$ from below. Of course this instantaneous equilibrium is immediately broken as the microprice drifts away from $S_t^+$ or $S_t^−$, possibly establishing new best bid and offer with non-vanishing size. This argument suggest that we could try to estimate the true value of the microprice from the best market prices and the corresponding posted volumes. A popular choice for estimating $M_t$ is

$$ M_t = \frac{V_t^- S_t^+ V_t^+ S_t^-}{V_t^+ V_t^-} $$

$\endgroup$
1
$\begingroup$

You would need to use Technical indicators like MacD, AROON, STOCH, etc. There are 100's of technical indicators which could be used to predict the price of the equity based on its past movements.

However, these indicators will help you in projecting the future price of an asset. The longer inputs you consider for these technical indicators, The longer future movement could be predicted. These indicators will suggest you on how much momentum in the price could be expected in the short/long run. But identifying the exact direction would be up to you. You would have probability of an asset going up or down. Based on the probability, you could act accordingly.

The big risks are. - Indicators have their own Limitations. *Evaluate for your positions periodically.
- You will not be able to consider the current/short term dynamics of the market. - Unexpected news and its impact on the asset owned. - Indicators doesn't consider any news and its impact to price. they assume everything is factored into the price of the asset. However it might not be true in real world. (Say, Enron or Merrill lynch)

A pessimistic example. Say, your indicator says, it would go up and you have plonked your money into the asset. After few days, bad news comes in from somewhere in the market, and the prices start to move in opposite direction. And your indicator will adapt to the recent changes in the price movements and it will adjust accordingly and will give you a different picture. You might argue that when invested, the graph was completely different, But current market dynamics force everything to change.

The only thing that's constant is Change. Change impacts everything in this current volatile market. you need to adapt yourself to the upcoming change and act accordingly.

I would suggest you to read about few technical indicators. Come back and read this question. Again read about few more indicators and come back to this question again. You would see a different picture and you would have few answers in your hand.

$\endgroup$
  • $\begingroup$ I think this question just answers how to compute or guess future stock price, and I think what the OP actually wants is how to guess minimum future stock price, which is quite a different (and more complicated) question. As @alex-c notes in his comment, it's more about assuming a known future distribution and then trying to find the minimum of that future distribution. $\endgroup$ – barrycarter Mar 22 '16 at 14:34
1
$\begingroup$

I like the maths here. However, I feel that the focus is off. You shouldn't really be wasting time trying to pick a "lowest possible price" or if you must then you should be trying to pick "all lowest possible prices". That is, a market correction. So why does the whole equity market move? You could have a look here or read reminiscences of a stock operator, but I think you're really asking for a formula to model the whole market.

To take some of the difficulty involved with accurately modelling the whole market, which may be possible some of the time, what about Warren Buffett's method? Buy into the stock of a company you can easily understand, and that has a nice and profitable business model that is making money, and hold it forever.

Ok, according to investopedia on the Buffett method, a key is to find a stock that is trading at a 25% discount to market value. Maybe that's what this question is all about. If so, as it says there, to work that out is all about calculating the intrinsic value of a business (get a value for corporate earnings, revenues, assets, brands, intangibles) and then seeing whether all that is actually reflected in the market capitalization.

Then we come back to where we started because to understand (and predict) why investors sometimes drive the market capitalization of a specific business down to levels where it trades at a big discount to its intrinsic value must be all about fears that the whole industry, or even the whole market, is going to crash. ie all about short term volatility... drat it can't be escaped!

In other words, investors sometimes drive the market capitalization of ALL companies down to below their intrinsic values, and you're into things like Tobin's Q ratio. That is an impressive theory but I don't think it's useful as a predictive indicator, and further, it's clear that after market corrections where everything is driven down in value, some companies recover their intrinsic value quicker than others...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.