# Relationship between ROE and IRR

In the textbook I read the following:

We can increase the present value of a share of common stock with a new investment only if $$ROE > r$$ , where $$r$$ is a discount rate (capitalization rate). If a new investment results in $$ROE < r$$, the price of the stock will decline even though earnings could be higher.

I have a trouble to understand the mathematical relationship bertween $$ROE$$ and $$r$$. In order to understand that I did the following operations. Since $$ROE=\frac{EPS}{P},$$ where $$EPS$$ is earnings per share, $$P$$ is stock price. From another hand, we have $$P=\frac{EPS}{r}+PVGO,$$ where $$PVGO$$ is present value of growth opportunities. Doing simple algebra the later yields: $$r=\frac{EPS}{P-PVGO}.$$ Does it mean that the difference between $$ROE$$ and $$r$$ is determined by $$PVGO$$? Please kindly explain the difference between those concepts.

• ROE is an accounting rate of return on the existing business, defined as $\frac{EPS}{BVE}$ where BVE is Book Value of Equity. $r$ is the expected return on the stock (determined in the stock market). They coincide in the hypothetical case of a no growth, or steady state company, which has no opportunities to expand its business by reinvesting earnings (hence it distributes all its earnings), but such companies are rare and so generally r and ROE are different. – Alex C Oct 9 '19 at 15:15
• @AlexC Thank you for very useful comment! So am i right that the difference between $ROE$ and $r$ is determined by $PVGO$? Is there any strict mathematical relationship between those two terms as I tried to derive in my posted question? – sane Oct 9 '19 at 16:43
• I could make up a problem where $r$ and $ROE$ have two arbitrary values, so I don't see the point in trying to find a mathematical relationship between two such conceptually different things. $ROE$ is associated with the business, $r$ has to do with the stock, i.e. how the stock market values the business (and its associated growth opportunities). – Alex C Oct 10 '19 at 15:52
• @AlexC Okay, thanks! – sane Oct 10 '19 at 15:58