If we have $f_X(x)=\left\{\begin{array} 11/4, 0\leq x\leq 1\\3/4, 1\leq x \leq 2 \\0, otherwise\end{array}\right.\quad\text{and} \; f_{Y|X}(y|x)\quad\left\{ \begin{array}{A}\frac{y-x} 2, x\leq y \leq x+2\\ 0, otherwise \end{array}\right.$ While observing the Random Variable Y=2.5. Find $\hat X_{MAP}$ and $\hat X_{LMS}$
Does anyone know how to solve this, I cannot find anywhere on the internet similar exercises, I would greatly appreciate the help!
My attempt at MAP $f_{Y|X}(y|x)f_X(x)= \frac{2.5-x}8$ $\frac d {dx} [\frac{-x}8]=-1/8$ MAP=0
Well, the maximisation for the MAP estimate in your answer is incorrect, because you're not accounting for the intervals over which $X,Y$ have density. Note that $$ f_{XY}(x,y) = \frac{y-x}{8}\left( \mathbf{1}\{x \in [0,1), y \in [x,x+2]\} + 3 \mathbf{1}\{x \in [1,2], y \in [x,x+2]\}\right).$$ Since $y \in [x,x+2] \iff x \in [y-2,y]$, this translates to $$ \frac{y-x}{8}( \mathbf{1}\{ \max(0,y-2) \le x < \min(1,y)\} + 3\mathbf{1}\{\max(1,y-2) \le x \le \min(2,y)\}).$$
Now, $$ \hat{x}_{\mathrm{MAP}}(2.5) = \arg\max_{x} f_{X|Y}(x|2.5) = \arg\max \frac{f_{XY}(x,2.5)}{f_Y(2.5)} = \arg\max f_{XY}(x,2.5),$$ as you correctly noted.But observe that if $y = 2.5,$ then for any $x < 0.5, f(x,0.5) = 0,$ so the MAP estimate must be at least $0.5$ (its in fact bigger). We can direclty work this out: $$ g(x) := f(x,2.5) = \frac{2.5-x}{8} \left(\mathbf{1}\{0.5\le x < 1\} + 3\mathbf{1}\{1 < x \le 2\} \right).$$ Notice that $g(x)$ falls over both $[0.5,1)$ and $[1,2]$, so the only possible maxima are at $0.5$ or $1$. By evaluating, $g(0.5) = 2/8,$ and $g(1) = 4.5/8, $ so $\hat{x}_{\mathrm{MAP}} = 1$.
For the least squares, i.e, for $\mathbb{E}[X|Y = 2.5],$ again note that this works out to $$ \int x f_{X|Y}(x|2.5) \mathrm{d}x = \int \frac{x g(x)}{f_Y(2.5)} \mathrm{d}x,$$ where recall that $$ f_Y(2.5) = \int_x f_{XY}(x,2.5)\mathrm{d}x.$$ Work out these integrals in whatever manner you find convenient (e.g., you don't need to worry about the $/8$s in either $f_{XY}$ or $g$ since they will cancel out), and take the ratio, which works out to about $1.26$.