Consider a random variable Y which has probability density function (pdf) defined by $f(y)=\frac{k\theta^k}{y^{k+1}}$ for$y\geq \theta$.

Question

Consider a random variable $Y$ which has probability density function (pdf) defined by $f(y)=\frac{k\theta^k}{y^{k+1}}$ for $y\geq \theta$.

How do I find the median of this function? I think it does not exist - how should I prove it? Can you please provide a detailed reason? Also there is a restriction in which k>2. Why is this necessary?

Answer 1

The median $m$ is defined as

$$\int_{-\infty}^m f(y)dy=\frac{1}{2} \implies \int_{\theta}^m\frac{k\theta^k}{y^{k+1}}dy=\left.-\left(\frac{\theta}{y}\right)^k \right\vert_\theta^m=-\left(\frac{\theta}{m}\right)^k+1=\frac{1}{2}.$$

We now have

$$\left(\frac{\theta}{m}\right)^k=\frac{1}{2}\implies m=\theta \cdot 2^{\frac{1}{k}},$$

which is fine since $2^{\frac{1}{k}}>1,$ for some positive, finite $k$.