Calculate Hazard Ratios [Survival Analysis] - YouTube

Hey guys, in this video

I'm going to do these two survival analysis problems and they're both on calculating a hazard ratio.

So basically for each problem we have two groups, group 1 and group 0, and we have the survival times

for these two groups and each survival time follows a certain probability distribution.

And we basically want to calculate,

given the probability distribution or whatever information, a ratio of the hazard functions for each of the two groups.

I'm gonna go ahead and do this problem first. For the first problem for group 1,

we're given that the survival times follow an exponential distribution where the parameter lambda is equal to 1 and for

group 0 we're given that the survival times follow an exponential

distribution with lambda equal to 2. The hazard functions for an exponential distribution

have a nice property to them but let's first assume that we don't know what the formula is for

calculating a hazard function from an exponential distribution. So we're just going to use the formula instead.

So a hazard function is equal to the PDF of a probability distribution divided by the

survival function. And the survival function in turn is equal to 1 minus the CDF of the probability distribution.

For group 1, we can calculate the hazard ratio as follows.

I'm just going to plug in the PDF of an exponential distribution,

which is lambda e to the negative lambda x over 1 minus the CDF of the exponential distribution,

which in turn is 1 minus e to the negative lambda x. This will simplify to

lambda times e to the negative x times lambda over

1 minus 1 plus e to the negative

lambda x.

One minus 1 is just 0 and then in the numerator and the denominator,

we can cancel out the e to the negative lambda x. This gives us just lambda.

So for any exponential distribution,

the hazard ratio is always just going to be equal to the parameter value of lambda.

So for the hazard ratio in group 1, lambda is equal to one.

The hazard ratio for group 1 is going to be equal to 1.

And for group 0,

the exponential distribution has lambda parameter equal to 2,

so the hazard ratio for group 0 is going to be equal to 2. And then the hazard ratio is just the ratio of

the hazard function for group 1 over the hazard function for group 2.

So for the first problem, that's just going to be a proportional hazards ratio of 1 over 2, and we're done with this first problem.

Then for this second problem,

we're basically going to do something similar. For group 1, we have that the survival times follow an exponential distribution

with lambda equal to 1 but for group 0 we're not given an exponential distribution

and we're also not given the actual probability distribution. We're given the survival function.

We have that the survival function for group 0 is equal to e to the negative x squared.

For group 1, we know the hazard function

for group 1 is just going to be equal to 1 because that's the same as how we did it for the last problem.

And now for group 0, we have to calculate from the survival function the hazard function.

One formula for doing that is that the hazard function is going to be equal to negative

the derivative of the log of the survival function.

Negative d log of the survival function dx will give you the hazard ratio for group 0.

So I'm going to do this step by step. I'm first going to take the natural log of the survival function.

When I do that, I'm just going to get negative x squared. Then

I'm going to take the derivative of the natural log of the survival function and that's going to give me negative 2x.

And then the final thing I have to do is take a negative on both sides.

The negative of the derivative is just going to give me 2x.

So in this case, the hazard ratio h1 over h0 is going to be equal to 1 over 2x and

also, this survival function or survival curve is a special case of the Weibull or Weibull distribution

which has two parameters

lambda and alpha.

Just as a general formula, the hazard ratio from a Weibull distribution

is alpha times lambda times x to the alpha minus 1.

So in this case, this is a solution for when the Weibull distribution

has lambda equal to 1 and alpha equal to 2. And we're done with this problem!

*chiptune music*