Using quotes in variables of a crontab job

I'm trying to get a past date inside a cron job using something like this. But the DATE variable is always empty.

DATE=date -d "$date -1 days -1 months" +%Y-%m
* * * * * /bin/echo "Date: $($DATE) Test" >> /tmp/crontab.log 2>&1

I know how tricky crontab is thanks to this question with good answers, but the problem here doesn't seem to be the % sign, because the code above works fine:

DATE=date -d @2147483647 +%Y-%m
* * * * * /bin/echo "Date: $($DATE) Test" >> /tmp/crontab.log 2>&1

But if I use quotes, then the job again can't get the date.

DATE=date -d"@2147483647" +%Y-%m
* * * * * /bin/echo "Date: $($DATE) Test" >> /tmp/crontab.log 2>&1

I tried replacing quotes with single quotes, double quotes, escaping the quotes, but none of this options solved the problem. And I need quotes to specify the "$date -1 days -1 months" part. Is there a way to do this in crontab without creating an external script?

CentOS 7 and crontab (cronie-1.4.11-23.el7.x86_64)

The issue has nothing to do with cron but with variable expansion within $(). The date command called in the subshell $() doesn't see a parameter -d "$date -1 days -1 months" but multiple params instead: -d "$date -1 days -1 months. You probably want to use $(eval $DATE) instead.