Leading P/E = (1-b)/ (r-g)
if payout ratio (1-b) = 100%, P/E = 1/ (r-g)
which means if when r approaching g, P/E ratio will rise very quickly.
But why@@ Any real meaning? Thanks.
Because all that growth in earnings is getting paid out to shareholders, making the equity more valuable (increase in price).
Or if r is decreasing towards g, then you are simply assigning less of a risk premium to the cash flow and therefore it’s more valuable.
These models assign value to stocks that return cash to shareholders. The higher the amount of NI that is returned to shareholders, the higher the stock is worth.
If ‘r’ approaches ‘g’, or better said, if ‘g’ approachs ‘r’, that means you’re getting more growth in compared to your required return. More growth means more NI and thus more dividends in the future and therefore a higher valuation.
I believe that it’s slightly more indirect than that.
If b approaches zero (so 1 – b approaches 1), then, to leave g unchanged, ROE has to increase. It’s the increase in ROE that causes the price of the stock to rise, not the fact that the earnings are being paid out as dividends.
This article that I wrote on justified ratios may help: http://financialexamhelp123.com/justified-ratios-price-multiples/.
Is that true in all cases or just this specific example? I think in this example (b = 0, g = 0) the stock would trade like a perpituity, so we couldn’t possibly leave g unchanged since g = b * ROE, g = 0 * ROE right?
Does ROE increasing mean the company has a higher ability to pay dividends to shareholders (and maintain this cash flow), and thus, a higher stock price today? I’m not sure i’m fully convinced that ROE is the key function to the stock price valuation, although I think i’m almost there!