Hello everyone,
I have tried to reproduce the isoprability countours (line) of a given dataset (see article link below) but I cannot manage to obtain the same results (using Matlab).
I am getting a bit desperate...
Here are the steps I have done:
0 Plot data
HR : Yaxis data
Q : Xaxis data
Color of symbols on plot indicate if stable (1) or unstable (0)
1 Calculate Betas
B0 (intercept) = 0.612
B1 (Q) = 1.104 plots on Xaxis
B2 (HR) = 0.422 plots on Yaxis
2 This allows to obtain equation:
z = B0 + B1x + B2y
z = 0.612 + 1.104Q  0.422HR
3 Knowing that:
logit(p) = ln(p / 1p) = B0+B1y+B2x = z [LOGIT function]
ln(p / 1p) = z [LOGIT function]
p / (1p) = exp(z) [ANTILOG of LOGIT function]
p = exp(z) (1p)
p = exp(z) – exp(z)p
p + exp(z)p = exp(z)
p [1+exp(z)] = exp(z)
p = exo(z) / [1+exp(z)]
The latest can be used to calculate the probability of making a successful prediction.
4 However, because we are interested in determining the isoprobability curve for various Q (B1) values, we need to isolate HR (B2) and include p in the equation.
From equations (0) & (3) :
(09) ln(p / 1p) = B0+B1x+B2y
(10) ln(p / 1p) – B0 – B1x = B2y
(11) [ln(p / 1p) – B0 – B1x]/B2 = y
5 For a given p, generate y outputs (HR) for various x values (Q) ranging from a to b.
The following two images are my curves (p = 5, 20, 40, 60, 80, 90, 95%) VS article's curves. Can't figure out why we don't end up with the same ? My 5 and 20% curves are very funky too!!
Did I miss something (well probably!!) ?
Thanks!!!
Click here for article
I have tried to reproduce the isoprability countours (line) of a given dataset (see article link below) but I cannot manage to obtain the same results (using Matlab).
I am getting a bit desperate...
Here are the steps I have done:
0 Plot data
HR : Yaxis data
Q : Xaxis data
Color of symbols on plot indicate if stable (1) or unstable (0)
1 Calculate Betas
B0 (intercept) = 0.612
B1 (Q) = 1.104 plots on Xaxis
B2 (HR) = 0.422 plots on Yaxis
2 This allows to obtain equation:
z = B0 + B1x + B2y
z = 0.612 + 1.104Q  0.422HR
3 Knowing that:
logit(p) = ln(p / 1p) = B0+B1y+B2x = z [LOGIT function]
ln(p / 1p) = z [LOGIT function]
p / (1p) = exp(z) [ANTILOG of LOGIT function]
p = exp(z) (1p)
p = exp(z) – exp(z)p
p + exp(z)p = exp(z)
p [1+exp(z)] = exp(z)
p = exo(z) / [1+exp(z)]
The latest can be used to calculate the probability of making a successful prediction.
4 However, because we are interested in determining the isoprobability curve for various Q (B1) values, we need to isolate HR (B2) and include p in the equation.
From equations (0) & (3) :
(09) ln(p / 1p) = B0+B1x+B2y
(10) ln(p / 1p) – B0 – B1x = B2y
(11) [ln(p / 1p) – B0 – B1x]/B2 = y
5 For a given p, generate y outputs (HR) for various x values (Q) ranging from a to b.
The following two images are my curves (p = 5, 20, 40, 60, 80, 90, 95%) VS article's curves. Can't figure out why we don't end up with the same ? My 5 and 20% curves are very funky too!!
Did I miss something (well probably!!) ?
Thanks!!!
Click here for article
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