I am trying to show that $$Var[X-I(X)] > Var[X-I_d(X)]$$ and I don't think this is true in general, so I would like to have some confirmation.
I am given $X \sim Unif(0,100)$ and $I(x)=kx, \quad k\in (0,1)$ while $I_d(x)= x-d, \quad x\in (d,100); 0 \text{ otherwise.}$
For simplicity let $Var[X] = \sigma^2$.
The LHS equals $(1-k)^2\sigma ^2$ and the RHS depends on the case of $x$.
If $x\in (0,d)$ then the variance is simply $\sigma^2$.
If $x \in (d,100)$ then the variance is $0$.
Because $(1-k)^2$ is a fraction the LHS seems to be less in this case . . .
Do you think I am right or am I calculating the variance of the RHS wrong?
I deeply appreciate your help.
You wrote: the RHS depends on the case of $x$. This is not true. The RHS is a number, it does not depend on any $x$ or $X$.
Let us rewrite the variable $X-I_d(X)$: $$ X-I_d(X)=X-(X-d)I_{\{X>d\}} = dI_{\{X>d\}}+ XI_{\{X\leq d\}} = \begin{cases} d, & \text{ if } X>d \cr X, & \text{ if } X\leq d \end{cases} $$ Find expected value of it: $$ \mathbb E[X-I_d(X)]= d\mathbb P(X>d)+ \mathbb E[X; X\leq d]=d\mathbb P(X>d) + \int_0^d xf_X(x)dx. $$ Substitute given pdf of $X$ and find the expectation. The second moment can also be found directly: $$ \mathbb E[(X-I_d(X))^2]= d^2\mathbb P(X>d)+ \mathbb E[X^2; X\leq d]=d\mathbb P(X>d) + \int_0^d x^2f_X(x)dx. $$ Then $\text{Var}(X-I_d(X)) = \mathbb E[(X-I_d(X))^2] - \left(\mathbb E[X-I_d(X)]\right)^2$.
You should get $\text{Var}(X-I_d(X)) = \frac{d^3}{300}-\frac{d^4}{4\cdot 100^2}$. It is not easy comparable to $(1-k)^2\sigma^2$.