# How to correct interest rate for inflation?

I'm looking for someone to double check my math and who can tell me if I have correctly accounted for inflation.

Say you hypothetically invest \$100 with 7% rate and 2% inflation. You would end up with \$107 at end of one year in future dollars. In today's dollars, that is worth 107/1.02=\$104.9, leading to an “effective rate” of 4.9%.

A formula I know of, which I'll call the "effective rate of return," is (1+interest rate)/(1+inflation)-1. Here it is (1+.07)/(1+.02)-1 =0.049. It correctly predicts the number above.

Some use the approximation of interest rate-inflation = .07-.02 =.05. I'm trying to interpret why it is only approximate. It is only approximate because it does not account for those 5 future dollars being worth less than \$5 current dollars. (Those \$5 future dollars are only worth 5/1.02=4.9 current dollars, as predicted by the “effective rate of return.")

So my question is, is the "effective rate of return" formula above the "correct" way to accommodate inflation for the purposes of calculating hypothetical future returns if one wishes to do it in today's dollars, and also if one wishes to account for inflation? I would be very thankful if someone can let me know if I had made mistakes, or if I am correct.

Edit: I think that the above formula works in all four combinations of positive/negative interest rates and positive/negative inflation rates. Above, I only presented the math for a positive interest rate with a positive inflation rate.

• `I'm trying to interpret why it is only approximate` - I'm not sure exactly what you want here, are you asking about the maths behind it? As you identify, the "true" effective rate is 4.9%, while the approximation gives 5%. It's quicker to calculate but slightly wrong - hence "only approximate". Sep 15, 2016 at 11:58
• I guess I'm attempting to interpret the approximation. Is it that the return of \$5 "in the future," which is approximately \$4.9 current dollars? Sep 15, 2016 at 12:34
• Side comment: "In today's dollars, that is worth \$104.9". Remember that depending on context, if you were attempting to calculate the value of an investment, it might be more accurate to say "In today's dollars, that is worth \$100." Consider: If you could invest money at 7%, with 2% inflation, would you rather have \$100 today, or \$104.9 next year? Both amounts are equivalent. This may seem like a nit-pic, but if you are talking in a finance context, then the total effective rate of return is lumped together to 'time value' money. Sep 15, 2016 at 12:57
• Interesting, I haven't been exposed to 'time value' money before. Can you maybe give a very brief summary? Sep 15, 2016 at 13:07
• @Grade'Eh'Bacon Well, "in today's dollars" is normally understood to mean, expressed in terms of the purchasing power of a dollar as of today, not the (usually lower) purchasing power that a dollar is expected to have in the future. What you're describing is "net present value", which is, basically, the amount I would have to invest today to get the specified amount of money at some future date. I've never heard that called "today's dollars". I suppose someone speaking loosely might say that, but I don't think it's accepted terminology.
– Jay
Sep 16, 2016 at 14:01

For 1 year horizon, the math is accurate. For a multi year horizon, compounding of interested rates earned and compounding due to inflation can take the effective rate much farther away from the simple math.

The answer will depend on what you want to use this for. If it is just to compare 2 simple investments, then yes. If it for more complex applications, then please use the compounded method.

For multiyear problem , effective interest rate will be ((1+interest rate )^N/ (1+inflation rate)^N) -1, where N i number of years

• Thank you for the reply, please let me know if I have understood it correctly. I feel as though I haven't So If I had 7% interest rate and 2% inflation rate for 10 years, expressed in today's dollars, you are saying that my \$100 would be become 100*(1+(1.07/1.02)^10-1))^10 = 11977 in today's dollars? That doesn't "seem" right, so my guess is that I haven't understood your answer. Sep 15, 2016 at 12:29
• Perhaps what you are saying is that my \$100 becomes 100*(1+(1.07/1.02)^10-1), which is \$161 and seems more reasonable. Sep 15, 2016 at 12:31
• I think I figured it out, if one plugs in my "effective rate" into the compound formula, 100*(1+(1.07/1.02)-1)^10 = 161, so I assume that the later of the two options is what you meant. Please clarify though if you would. Sep 15, 2016 at 12:36
• In terms on money that flows into your hand : 100*(1.07)^10=196. Cost of something which is \$100 today, after 10 years , will become 100*(1.02)^10=121. Now, the answer that you are looking for is, what will the 100 dollars today, effectively amount for 10 years later, you need to find effective intreset rate R and then use 100* (1+R)^10, where R will be (1.07/1.02)^10-1= 1.613-1=0.613. This is same as (196/121)-1 Sep 15, 2016 at 13:14
• Yes, and I understand that to be 197 future dollars. If I wish to understand how much that is worth in today's dollars, it would then be 197/(1.02^10)=\$161, as calculated in above. Thanks! Sep 15, 2016 at 13:18

Yes. The math is right. As is your explanation. But, here's the thing to consider - is the 2% rounded, or does it have an accuracy of 2.0%? You can't really take two numbers rounded to integers, and end with an extra decimal place of accuracy.

• Fair point, I was playing loose with significant digits. Sep 15, 2016 at 12:31

It is only approximate, because the product of two numbers close to 1, is very close to the sum of those numbers. (Likewise, dividing two numbers can be approximated by subtracting them.)

For example, `1.01 * 1.01 = 1.0201`, which is very close to the answer you get if you simply add them, `1.02`.

Additionally, people typically find it easier to add/subtract numbers than to multiply/divide. (This is doubly true if the numbers are stated as percentages, because you can skip the steps of converting from a percentage into a decimal and back again.)

So, it is technically wrong to say that two independent 1% gains will result in a 2% overall gain (the correct figure is 2.01%) - but it is much quicker/easier to calculate. And if the original figures have any uncertainty to them (which is the case with your yield and inflation predictions) then this will probably dwarf the slight inaccuracy in the calculation anyway.

The math here is correct for the problem \$100 now versus \$107 a year from now. Note however that interest usually isn't calculated that way. Instead, interest compounds monthly or daily. If you take a 7% interest rate compounded monthly, you actually would have \$100 now versus \$107.23 (rounded up) a year from now.

``````100 * (1 + .07 / 12)^12 = 107.229...
(107.229... / 1.02) - 1 = 105.126...
``````

Again, that's a 7% annual nominal rate compounded monthly giving an effective annual percentage rate of charge (effective APR) of 7.23% (rounded up).

Of course, 7% would be an incredible interest rate at the moment, so perhaps you are using an estimated rate of return for securities instead. That would make more sense with your original numbers.

You can get inflation rate numbers that run at whatever time period you wish. The Consumer Price Index (CPI) and its variants are calculated monthly, and you can use them to calculate a rate for any month to any other month.

We can make this even more complicated by introducing payments. As is, we are comparing money now to money a year from now. But in many circumstances we'd have a lump sum now and monthly payments back or monthly deposits to a total a year from now. In those cases, you'd also have interest and inflation on the monthly amounts.

As I said at the beginning, your math is correct for the simple case of \$100 now and \$107 a year from now assuming 2% inflation. It's just that that is itself potentially a simplification of a more complicated scenario. Don't get so focused on the dime there that you miss the \$0.22 from compounding or whatever other specific issues exist in the real scenario.

• Interesting. Can you post the "effective rate of return" formula that can be used for when compounding n times per year? For n = 1, it should reduce to the equation in my above post. Oct 6, 2016 at 18:59

Some use the approximation of interest rate-inflation = .07-.02 =.05. I'm trying to interpret why it is only approximate.

The name of the approximation is Fisher equation

I belive this section of Wikipedia is sufficient.

https://en.wikipedia.org/wiki/Fisher_equation#Derivation • Interesting, good to know I guess. I'll just use the exact result though I suppose because it's easy to code. Oct 6, 2016 at 19:01