# Total Variation Distance Uniform Distribution

Hello I am trying to solve the following but the answer is wrong and I cant seem to see my mistake.

Question :

Find the total variation distance between

P = Unif([0,s]) and Q = Unif([0,t]) where 0 < s < t

Calc TV(P,Q)

I applied the formula which is (1/2) ( Integral | ftheta(x) - ftheta'(x) |)

So I got 1/s for P and 1/t for Q.

My TV would be (1/2) * Integral (1/s - 1/t)

Can you tell me where I am getting this wrong or how i should go from there. I am getting confused on the way forward.

Thank you

BS

The total variation is $$TV(\theta,\theta') = \int_E |f_\theta(x)-f_{\theta'}(x)|$$ where $$E$$ is the support. For this question, the support is split into two regions $$[0,s]$$ and $$[s,t]$$.
$$TV(P,Q) = \frac{1}{2} \left(\int_0^s \left|\frac{1}{s} - \frac{1}{t}\right|dx + \int_s^t \left|0 - \frac{1}{t}\right|dx\right) = \frac{t-s}{t}$$.