Is there a statistical way in which I can age-match these groups even though children in Group B are systematically younger than children in Group A?

I've been introduced very briefly to propensity scores, but I'm just not sure.

- Thread starter Barnzilla
- Start date

Is there a statistical way in which I can age-match these groups even though children in Group B are systematically younger than children in Group A?

I've been introduced very briefly to propensity scores, but I'm just not sure.

Suppose your ages are continuous variables (this might be appropriate if you have a lot of birthdays), a one-to-one correspondence can be created easily:

x - age G1, 5 <= x <= 7

y - age G2, 2 <= y <= 5.

Let f(x) = y. We seek f(5) = 2, and f(7) = 5, as well as f(x) being linear (i.e. injective).

f(5) = 2 = a*5 + b -> b = 2 - a*5

f(7) = 5 = a*7 + b -> b = 5 - a*7

2 - a*5 = 5 - a*7 -> a = 2*a = 3 ->

Now f(x) = 3/2x - 11/2, so that f(5) = (3/2)*5 - 11/2 = 2, and f(7) = (3/2)*7 - 11/2 = 5. f(x) effectively maps the interval [5, 7] to [2, 5]. If you want to map [2, 5] to [5, 7], use h(y) = (y + 11/2)*(2/3).

I'd only use this approach if you can accurately pair ages in group 1 with ages in group 2.

Hope it provides some insight

Suppose your ages are continuous variables (this might be appropriate if you have a lot of birthdays), a one-to-one correspondence can be created easily:

x - age G1, 5 <= x <= 7

y - age G2, 2 <= y <= 5.

Let f(x) = y. We seek f(5) = 2, and f(7) = 5, as well as f(x) being linear (i.e. injective).**f(x) = ax + b**. Solving for a, b:

f(5) = 2 = a*5 + b -> b = 2 - a*5

f(7) = 5 = a*7 + b -> b = 5 - a*7

2 - a*5 = 5 - a*7 -> a = 2*a = 3 ->**a = 3/2** -> **b = 2 - (3/2)*5 = -11/2**

Now f(x) = 3/2x - 11/2, so that f(5) = (3/2)*5 - 11/2 = 2, and f(7) = (3/2)*7 - 11/2 = 5. f(x) effectively maps the interval [5, 7] to [2, 5]. If you want to map [2, 5] to [5, 7], use h(y) = (y + 11/2)*(2/3).

I'd only use this approach if you can accurately pair ages in group 1 with ages in group 2.

Hope it provides some insight

x - age G1, 5 <= x <= 7

y - age G2, 2 <= y <= 5.

Let f(x) = y. We seek f(5) = 2, and f(7) = 5, as well as f(x) being linear (i.e. injective).

f(5) = 2 = a*5 + b -> b = 2 - a*5

f(7) = 5 = a*7 + b -> b = 5 - a*7

2 - a*5 = 5 - a*7 -> a = 2*a = 3 ->

Now f(x) = 3/2x - 11/2, so that f(5) = (3/2)*5 - 11/2 = 2, and f(7) = (3/2)*7 - 11/2 = 5. f(x) effectively maps the interval [5, 7] to [2, 5]. If you want to map [2, 5] to [5, 7], use h(y) = (y + 11/2)*(2/3).

I'd only use this approach if you can accurately pair ages in group 1 with ages in group 2.

Hope it provides some insight

I think at the bottom line, age difference is serious issue in most situations, and there might not be much to do to make it "as if there isn't difference." But I might try to standardise the dependent variables in each group, and use the group-wise standardised numbers to do the rest of analysis.