Linear estimation of an exponential distribution QUESTION
We have $Y \sim \mathrm{Exp}(1/6)$. We define $T = e^{−4Y}$
Calculate the best linear estimator of $T$ according to $Y$
ANSWER
Ok it sounds pretty simple at first
I got my $f_Y(y)=\frac{1}{6} \cdot e^{-\frac{y}{6}}$.
I am pretty sure the equation for an estimation is $g(Y)=E[T\mid Y]$
First i need the density function of $T$ given $Y$ : $h(t\mid y)=\frac{f(y,t)}{f_Y(y)}$. Then, $E(T\mid Y)=\int t\cdot h(t\mid y) \, dt$
I have $f_Y(y)$ , but how i find $f(y,t)$
Thank you in advance
 A: For now I'll guess that by "linear" you mean an estimator of the form $a+bY$ where $a$ and $b$ are constants, i.e. are not random.  (But maybe one could also mean just $bY$, without the $a$?)
"Best" is sometimes taken to mean minimizing the expected square of the residual $T - (a+bY)$, i.e. choose $a$ and $b$ to make
$$
\operatorname{E}\Big( ( T - (a+bY) )^2 \Big)
$$
as small as possible.  So compute the expected value:
\begin{align}
& \operatorname{E}\Big( ( T - (a+bY) )^2 \Big) = \int_0^\infty \left( e^{-4y} - (a+by) \right)^2 e^{-y/6} \, \frac{dy} 6 \\[10pt]
= {} & \int_0^\infty \left(  e^{-8y} +a^2 + b^2y^2 - 2ae^{-4y} - 2by e^{-4y} + 2aby \right) e^{-y/6} \, \frac{dy} 6.
\end{align}
To evaluate this, recall that
\begin{align}
\int_0^\infty e^{-ry}\,dy & = \frac 1 r \\[10pt]
\int_0^\infty y e^{-ry}\,dy & = \frac 1 {r^2} \\[10pt]
\int_0^\infty y^2 e^{-ry} \, dy & = \frac 2 {r^3}
\end{align}
When you're done, you'll have a quadratic function of $a$ and $b$.  You need to find the value of $(a,b)$ that minimizes it.
