Can I give a discount on fees for referrals?
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I am writing a 12 mo business plan for a new firm I am looking at. Before including my idea of reducing fees for each referral I get I wanted to make sure compliance and FINRA will allow it. Anyone know? Can I do it by dollar amount say $500 per every referral that becomes a client, or like 20 basis points?
Does anyone currently do this in their own practice?
Thank for your help guys.
I am in no way qualified to answer this...
But I have been reading over numerous ADV Part 2's recently so I know that you can give discount for fees based solely on your discretion.
I don't do fee based accounts, but I'm pretty sure that's not gonna fly. But don't take my word for it.
Yeah, FINRA won't even warn you about that one. That's a real quick way out of the industry.
Sounds a lot like that is paying fees/commissions indirectly to unlicensed individuals. And, that's a no, no, my friend. Why not just give gifts for referrals as most others do? But, be careful on the dollar amount of gifts. There are limits on gifts as well.
Are you talking about discounting the new account or paying the guy who refered you the account?
My understanding is that you can discount your own fees for the client but you can't give a monetary kick back to the source of the referral, ie "Thanks for the referral Bob, here's $500 for the trouble".
You can only give your own clients $100 a year for things like dinner, basketball games, ect - unless you attend with him in which case there's no limit.
But if you say "Hey, here's two tickets, have fun" - they better not be more then $100.
-PF
PushForward- My understanding is the same at you. The only way I can pay someone in cash or gifts over 100/yr is to have them a licensed solicitor and have it disclosed.
My idea, and I know it falls in the gray area of compliance, can I discount a CURRENT clients FEE if they refer me another client. I thought when I was getting my 66 they said I could discount my fee for any reason, of course depending on the companies policy also.