I want to the derive the compound PDF of a triangular and uniform distribution which looks like this:

To to this I have simply added the PDFs for the rectangular and triangular parts, over the range $[n,N].$

A triangular distribution, with these bounds, has the following PDF:

$$f(x) = \frac{2(N-x)}{(N-n)^2}$$

The scaled uniform distribution has the following PDF:

$$g(x) = \frac{1}{N-n}$$

Then (I believe), the compound distribution is simply:

$$h(x) := f(x) + g(x) = \frac{3N -2x -n}{(N-n)^2}$$

However, I do get a bit confused here, since this distribution needs to be normalised, which is simply done as so:

$$h_{\text{norm}}(x) = \frac{1}{\int_x h(x)} h(x)$$

Does this seem reasonable, or am I wildly off-chart here?

This is a related question but it seems very complicated, for what should be quite simple.

I want to the derive the PDF which looks like the sum of a triangular and uniform distribution which looks like this:

To do this I have simply added the PDFs for the rectangular and triangular parts, over the range $[n,N].$

A triangular distribution, with these bounds, has the following PDF:

$$f(x) = \frac{2(N-x)}{(N-n)^2}$$

The scaled uniform distribution has the following PDF:

$$g(x) = \frac{1}{N-n}$$

Then (I believe), the compound distribution is simply:

$$h(x) := f(x) + g(x) = \frac{3N -2x -n}{(N-n)^2}$$

However, I do get a bit confused here, since this distribution needs to be normalised, which is simply done as so:

$$h_{\text{norm}}(x) = \frac{1}{\int_x h(x)} h(x)$$

Does this seem reasonable, or am I wildly off-chart here?

This is a related question but it seems very complicated, for what should be quite simple.