Suppose we are given a probability function, P (x^T (Y-z)≥0) , where ‘x’ is a vector, ‘Y’ is a random variable and ‘z’ is a known value.
Now, suppose, we make a non-singular transformation w=Ax, where ‘w’ is a vector and ‘A’ is a non-singular matrix.
Now let us define, another probability function, P (w^T (Y-z)≥0) .
My question now is – does the “infimum” of the above-given probability function remain same after undergoing the non-singular transformation ? If so, how is this possible ? Can anyone please explain ? Thanks in advance.
P.S. : x^T and w^T means "x transpose" and "w transpose" respectively. Pardon me for this. I couldn't format this, actually.
Assuming that $x$, $A$, and $z$ are fixed, define the sets
$$ T_1 = \{y\,:\, x^T(y - z) \geq 0\} $$
and
$$ T_2 = \{y\,:\, (Ax)^T(y - z) \geq 0\} $$
Then you can restate the question in terms of $\mathrm{P}(Y \in T_1)$ and $\mathrm{P}(Y \in T_2)$. Clearly, we have that
$$ \inf \mathrm{P}(Y \in T_1) \leq \inf \mathrm{P}(Y \in T_2) ~~\text{if}~~ T_2 \subset T_1 $$
and
$$ \inf \mathrm{P}(Y \in T_2) \leq \inf \mathrm{P}(Y \in T_1) ~~\text{if}~~ T_1 \subset T_2 $$
Since $z$ belongs to both sets, the only other case is when $T_1\cap T_2 \neq \emptyset$ and $T_1 \not\subset T_2$ and $T_2 \not\subset T_1$. However, it is difficult to say anything about the relationship between $\inf \mathrm{P}(Y \in T_1)$ and $\inf \mathrm{P}(Y \in T_2)$ in this case without more information about $\mathrm{P}$ and $A$.
I'm not sure if this is helpful, but if you define $X = \mathrm{span}\{x\}$ and $W = \mathrm{span}\{Ax\}$, then it follows that
$$ T_1 = \{z + \alpha x\,:\, \alpha > 0\}\cup\{z + X^\perp\} \quad\text{and}\quad T_2 = \{z + \alpha Ax\,:\,\alpha > 0\}\cup\{z + W^\perp\} $$