How can I prove that the hazard function for $Y$ is the following $$h_Y(y)=\frac{1}{k}h_V\left(\frac{y}{k}\right)$$ when I am told that $V$ has a probability density function $f_V(v), v>0$, hazard function $h_V(v)$ and the transformation to be used is $Y=kV$ where $k$ is positive constant?

I know I have to calculate the cdf, survival function and then the rate of these two suppose to be the hazard function, but where do I start when I am only told that $V$ has pdf and no pdf is defined?


You do not need the exact form of the pdf of $V$, but only that it exists and the fact that you can express the cdf $F_Y(y)$ of $Y$ in terms of the cdf $F_V(v)$ of V $$F_Y(y)=P(Y\le y)=P(kV\le y)\overset{k>0}=P\left(V\le \frac{y}{k}\right)=F_V\left(\frac{y}{k}\right)$$ Hence, since you know that $V$ has a pdf you can differentiate the above expression (with respect to $y$) and express also the pdf $f_Y(y)$ of $Y$ in terms of the given pdf $f_V(v)$ of $V$: $$f_Y(y)=\frac{d}{dy}F_Y(y)=\frac{d}{dy}F_V\left(\frac{y}{k}\right)=f_V\left(\frac{y}{k}\right)\frac{d}{dy}\left(\frac{y}{k}\right)=\frac1kf_V\left(\frac{y}{k}\right)$$ Hence $$h_Y(y)=\frac{f_y(y)}{1-F_Y(y)}=\frac{\frac1kf_V\left(\frac{y}{k}\right)}{1-F_V\left(\frac{y}{k}\right)}=\frac1k\frac{f_V\left(\frac{y}{k}\right)}{1-F_V\left(\frac{y}{k}\right)}=\frac1kh_V\left(\frac{y}{k}\right)$$

  • 1
    $\begingroup$ This was very helpful. I knew I could get though with knowing that pdf exists but couldn't think of an initial step. Your first formula made it very clear. Thank you. $\endgroup$ – lovetimberland Oct 13 '16 at 8:45

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