So let's say that we have a market-cap weighted ETF (free-floating shares) that tracks an index comprised of stocks A and B. Stock A has a share price of $$10 and there are 100 outstanding shares and stock B has a share price of $50 and there are 5 outstanding shares. Let's add that the divisor of the index this ETF tracks is 100. The goal of the ETF at inception is to provide 2:1 long (Bull) leverage.
First question: The ETF has 2500 dollars worth of exposure (because of 2:1 leverage), and with the divisor of 100, would each share of the fund have a price of 12.5 or 25 at the time of ETF origination?
Second question: Does it make sense to say that there are 500 available shares of the ETF? Would we then set the price of each of the shares to $5 (2500/5) or would that make the market cap of the etf greater than the market cap of the stocks it attempts to track (500*12.5). What is the maximum number of shares that this fund can create?
Third question: If this is a 2:1 leveraged etf - let's take a look at how it rebalances after a day of trading.
Assuming the price of the etf is 12.5, we buy 8 shares for total assets of 100. The etf, however, gives us 200 dollars worth of exposure (16 shares of A and 0.8 shares of B). Stock A rallies 5%, and stock B rallies 10%. Our exposure is now: 168 (from A) + 44 (From B) = 212. We now have $112 in assets, which means our shares increased in price from 12.5 to 14 dollars.
The fund now needs to buy an additional $12 (112*2=224-212) dollars worth of stocks to rebalance at the end of the day. It needs to buy these according to the new market cap weightings: 10.5*100+55*5 = 1325. Stock A will take up (1050/1325)% of this purchase (about 79%), or 9.48 dollars which is 9.48/10.5=-.902 shares of stock A. We buy 2.52 dollars of stock B, which is 2.52/55=0.045 shares of stock B. Thus we have exposure to:
16.902 shares of stock A and 0.845 shares of stock B. Is that correct?
In general I'm pretty loose on my understanding of this topic. Any information or resources are greatly appreciated.