I would like to ask:
Let assume that the mid-price now is m0,
then a trade happens and the new mid-price is m1. Is it possible to have
m1 = m0 or it should be always different?
Thank you in advance.
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If you define the mid price as the average of the best bid and best ask price the answer is no. If there are bids for 10 lots @ 10.00 and asks for 10 lots @ 10.10 markets orders smaller than 10 lots will not affect the mid price. The mid is 10.05 before and after.
This also implies that no trade is necessary to change the price. If the best bid or ask changes, the result of $$\frac{\mathrm{best\,bid} + \mathrm{best\,ask}}{2}$$ will also change.
answered Mar 7 '17 at 13:34
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1$\begingroup$ Also, the mid can change without any trade at all. if you assume that there is only one guy standing at the best bid or best ask and he cancels his order, the mid would change. $\endgroup$ – nimbus3000 Mar 7 '17 at 13:38
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$\begingroup$ Excellent point! I've added it to my answer. $\endgroup$ – Bob Jansen♦ Mar 7 '17 at 13:41
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$\begingroup$ It's also possible, in markets with quantized time (e.g. many currency ECNs), for a trade to happen that takes out one or more price levels without changing the mid price. $\endgroup$ – Chris Taylor Mar 7 '17 at 13:43
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$\begingroup$ Off late many exchanges have hidden liquidity at midpoint. nasdaqomx.com/transactions/trading/midpoint . If your order is configured to trade against those shares, the trade will not change the mid. Although it would be mostly applicable for institutional flow. $\endgroup$ – Swagato Acharjee Mar 9 '17 at 18:29
