I have been given the mean, median and mode of a function and have to find the probability density function.
mean: $\gamma - \beta\Gamma_1$
median: $\gamma-\beta(ln2)^{1/\delta}$
mode: $\gamma-\beta(1-1/\delta)^{1/\delta}$
I am also given that $$\Gamma_k=\Gamma(1+k/\delta)$$ $$\Gamma(z)=\int_0^\infty t^{z-1}dt$$ $$ -\infty<x<\gamma, \beta>0, \gamma>0 $$
Now I understand how to calculate the mean, mode and median when given a probability density function. However I'm struggling to go backwards. I initially tried to "reverse" the process by differentiating the mean or median however I know this is skipping the substitution over the given limit.
I then looked for patterns with known distributions and realised they are from Weibull distribution however $\gamma-$. Does this mean essentially this is a typical Weibull distribution however shifted by $\gamma$ and therefore the pdf will be $\gamma-Weibull pdf"$
Given just the mean, median, mode and maximum, I would expect there to be an infinite number of possible distributions. But you seem to be correct that the obvious one is related to the Weibull distribution
What you are saying is that $W$ is a Weibull random variable with distribution scale parameter $\beta$ and shape parameter $\delta$ then one possibility is that the random variable you are looking could be $X=\gamma - W$
This Weibull distribution would have density $$f_W(x)=\begin{cases} \frac{\delta}{\beta}\left(\frac{x}{\beta}\right)^{\delta-1}e^{-(x/\beta)^\delta} & x\geq0\\ 0 & x<0\end{cases}$$
so if $X=\gamma - W$ then $X$ would have density $$f_X(x)=\begin{cases} \frac{\delta}{\beta}\left(\frac{\gamma-x}{\beta}\right)^{\delta-1}e^{-((\gamma-x)/\beta)^\delta} & x \leq \gamma\\ 0 & x>\gamma\end{cases}$$