Left truncated normal distribution
MDP
I am interested in the following problem.
Suppose we have a left truncated normal distribution (one-dimension)
for which the expectation and the variance are known. What are the
mean and variance of the corresponding normal distribution ?
I know it is straightforward to compute the expectation and variance
of a LTN starting from a known normal distribution and a lower bound
but I can't see how I can do the reverse operation in a quick way. Of
course I can think about numerical optimization but I really need a
fast answer because the operation will be repeated many many times.
Any idea ?
Michael
Ray Koopman
You know *only* the mean and variance of the truncated distribution?
You know neither the cutoff nor the proportion that has been cut off?
Sounds pretty hopeless.
Greg Heath
I only get two equations for the three unknowns m0,s0 and (T-m0)/s0.
You need another piece of information. For example,
the lower bound T.
Hope this helps.
Greg
MDP
Ok, I forgot to mention it. Yes the cut off value is known as well, it
is zero.
Ray Koopman
M = mean of the entire distribution
S = s.d. of the entire distribution
c = cutoff point; c = 0 in the case under consideration.
m = mean of the truncated distribution
s = s.d. of the truncated distribution
m > M and m > c, because the lower tail has been cut off.
s < S, whichever tail has been cut off.
f = standard normal pdf
P = standard normal cdf = lower tail area
Q = 1-P = upper tail area
The basic equations are
m-M f[(c-M)/S]
--- = ---------- (1)
S Q[(c-M)/S]
and
s^2 m-M m-M c-M
--- = 1 - (---)*((---)-(---)). (2)
S^2 S S S
Setting c = 0 and letting f/P = g, (1) becomes
m = M + S*g[M/S], (3)
and (2) becomes
s^2 = S^2 - (m-M)*m. (4)
Substituting (3) into (4) gives
s^2 = S^2 - (S*g[M/S])*(M + S*g[M/S]). (5)
Let M/S = T. Then (3) becomes
m = S * (T + g[T]) (6)
and (5) becomes
s^2 = S^2 * (1 - g[T]*(T + g[T])). (7)
Dividing (6) by the square root of (7) gives
T + g[T]
m/s = --------------------- = h[T]. (8)
sqrt[1-g[T]*(T+g[T])]
There is no obvious (to me) way to invert h and write T as an
explicit function of m/s. But h is smooth and concave-up, with
asymptotes 1 and T, so it should be solvable numerically without
any special problems as long as T doesn't get too negative.
In any case, once you have T you can get S from (6) and then M = S*T.
Greg Heath
OK. Now you have two equations and two unknowns.
Why don't you derive the equations and post them?
Hope this helps.
Greg
MDP
Many thanks for your answers !
Michael