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Wrong estimate in query plan

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In a complex query I have query I noticed that the planner made a bad estimate for a join between two tables in the query (I made sure statistics were up to date).

The join was on a single column. In the first table are about 985 rows. All rows except one have a NULL value in  the join column the one value in that column is 1. The other table has 237240 rows and all rows have a 1 in the join column. This column cannot contain NULL values. There is a foreign key constraint between the join columns.

In pgAdmin I had a look at the statistices for the two columns.

Column in first table
Null Fraction 0.998985
Average Width 4
Distinct Values -1
Most Common Values
Most Common Frequencies
Histogram Bounds
Correlation

What I noticed is that are no most common values mentioned ofcourse the value 1 only occurs once in the column but as all other values are NULL you could argue it is a common value.


Column in second table:
Null Fraction 0
Average Width 4
Distinct Values 1
Most Common Values {1}
Most Common Frequencies {1}
Histogram Bounds
Correlation 1

Looks fine :)


Relevant part of AXPLAIN ANALYZE output
'                                ->  Hash Join  (cost=40.16..6471.62 rows=241 width=58) (actual time=0.486..102.979 rows=237240 loops=1)'
'                                      Hash Cond: (kb.filiaal_id = fil.filiaal)'
'                                      ->  Seq Scan on kassabon kb  (cost=0.00..5539.40 rows=237240 width=42) (actual time=0.036..28.562 rows=237240 loops=1)'
'                                      ->  Hash  (cost=27.85..27.85 rows=985 width=20) (actual time=0.434..0.434 rows=1 loops=1)'
'                                            Buckets: 1024  Batches: 1  Memory Usage: 1kB'
'                                            ->  Seq Scan on relatie fil  (cost=0.00..27.85 rows=985 width=20) (actual time=0.003..0.382 rows=985 loops=1)'

Notice how it expects 240 rows but gets 237240.

I wondered if this should be reported as a bug? It goes wrong I think because of the statistics of the column in the first table give no information about the values present except the NULL values.
Another thought I had was that the planner could have known there was a matching row in the first table for each row in the second table because there is a foreign key constraint between the two.


Regards, Eelke

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