Variance of $1/Y$ where $Y$~Gamma I need help computing for $Var(\frac{1}{Y})$ where $Y$~Gamma(r,λ)
I know that $$Var(\frac{1}{Y}) = E(\frac{1}{Y}^2) - [E(\frac{1}{Y})]^2$$
I have computed for:
$$E(\frac{1}{Y}) = \frac{λ}{r-1}$$
Here is my solution so far for $E(\frac{1}{Y^2})$:
$$ = \int_0^∞ \frac{1}{y^2} \frac{λ^r}{Γ(r)} y^{r-1} e^{-λy} dy$$
$$= \frac{λ^r}{Γ(r)} \int_0^∞ y^{(r-1)-2} e^{-λy} dy$$
$$= \frac{λ^r}{Γ(r)} \int_0^∞ y^{(r-2)-1} e^{-λy} dy$$
$$= \frac{λ^r}{Γ(r)} \int_0^∞ \frac{Γ(r-2)}{λ^{r-2}}\frac{λ^{r-2}}{Γ(r-2)}y^{(r-2)-1} e^{-λy} dy$$
$$= \frac{λ^r}{Γ(r)} \frac{Γ(r-2)}{λ^{r-2}} \int_0^∞ \frac{λ^{r-2}}{Γ(r-2)}y^{(r-2)-1} e^{-λy} dy$$
The integral is the PDF of Gamma (r-2,λ) so
$$= \frac{λ^r}{Γ(r)} \frac{Γ(r-2)}{λ^{r-2}} (1)$$
$$= \frac{λ^r}{λ^{r-2}} \frac{Γ(r-2)}{Γ(r)}$$
$$= λ^2 \frac{Γ(r-2)}{Γ(r)}$$
I am not sure where to go from here.
 A: It is worth observing that if $$Y \sim \operatorname{Gamma}(r, \lambda)$$ where $r$ is a shape parameter and $\lambda$ is a rate parameter, then $W = 1/Y$ is called an inverse gamma random variable, which has density
$$f_{W}(w) = f_Y(1/w) \cdot \frac{1}{w^2} = \frac{\lambda^r (1/w)^{r-1} e^{-\lambda/w}}{w^2 \Gamma(r)} = \frac{\lambda^r e^{-\lambda/w}}{w^{r+1} \Gamma(r)}, \quad w > 0.$$
If we accept that $$\int_{w = 0}^\infty f_W(w) \, dw = 1,$$ that is to say, $f_W$ is a density for any choice of parameters $r, \lambda > 0$, then it is easy to compute the moments.  Let $0 \le k < r$.  Then
$$\begin{align}
\operatorname{E}[W^k] 
&= \int_{w=0}^\infty w^k f_W(w) \, dw \\
&= \frac{\lambda^r}{\Gamma(r)} \int_{w=0}^\infty \frac{e^{-\lambda/w}}{w^{r-k+1}} \, dw \\
&= \frac{\lambda^r \Gamma(r-k)}{\lambda^{r-k} \Gamma(r)} \int_{w=0}^\infty \frac{\lambda^{r-k} e^{-\lambda/w}}{w^{r-k+1}\Gamma(r-k)} \, dw \\
&= \lambda^k \frac{\Gamma(r-k)}{\Gamma(r)},
\end{align}$$
where in the last step, the integrand is the density of an inverse gamma distribution with shape $r-k$ and scale $\lambda$.  Then for $k = 1$ and $k = 2$, we obtain
$$\operatorname{E}[W] = \frac{\lambda}{r-1}, \\ \operatorname{E}[W^2] = \frac{\lambda^2}{(r-1)(r-2)},$$ and $$\operatorname{Var}[W] = \frac{\lambda^2}{(r-1)^2(r-2)},$$ provided that $r > 2$.
Your solution is basically correct, but you just need to recognize that $$\frac{\Gamma(r-2)}{\Gamma(r)} = \frac{\Gamma(r-2)}{(r-1)(r-2)\Gamma(r-2)}$$ since $z \Gamma(z) = \Gamma(z+1)$.
A: Also observe that
If $Y\sim Gamma$ then $\frac{1}{Y}\sim$ Inverse Gamma thus mean and variance are known indicators
