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I generally use Google throughout the day to track the price of stocks. I understand there can be minor discrepancies between sources based on the time the data is collected and how it is displayed but the discrepancy pictured below I cannot seem to understand.

enter image description here

On the left is the price ticker for the day from Google and as you can see there is a large spike around 10:30AM. I assumed this was from a large volume market order being filled. To check I went to Fidelity's site but found no such volume or spike on their chart.

What is the cause of this?

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    I suspect it's just a bad data point. What's odd is that the spike does not appear on Google Finance, but just in the search results. – D Stanley Mar 21 '17 at 20:39
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enter image description here

This is from Google Finance right now.

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    The image in the question is from Google's search results, not Google Finance, although it's odd that they're different. – D Stanley Mar 21 '17 at 20:38
  • Hmm.. Odd then I guess. Depending on how the search result table is made I guess it could be possible that just 1 share sold at the price listed at the correct time could have created the data point.. – KDecker Mar 21 '17 at 20:50
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    I get that this is different than the search result, it's probably some search caching mechanism that had an error. But clearly Google Finance is tracking properly, so unless one of the users here is a Google engineer working on search result parsing there isn't much to say. – quid Mar 21 '17 at 20:51
  • @KDecker No, one share could not be traded outside of the bid/ask range. The only way that were possible is if there were no bids lower than 16.45, but within 5 minutes the price is back down to 15.44. I highly suspect it's a data glitch - Looking on ETrade, which has a 1-minute resolution, that spike is not present. – D Stanley Mar 21 '17 at 21:21
  • Looking at Google Finance's graph for today you can even see an issue (but in reverse). Around 12:45 this is a negative price spike that doesn't seem to actually have a point value on the graph. – KDecker Mar 22 '17 at 16:58

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