Question

Problem 4 Let X be a discrete random variable with probability mass function fx(x), and let t be a function. Define Y = t(X): that is, Y is the randon variable obtained by applying the function t to the value of X Transforming a random variable in this way is frequently done in statistics. In what follows, let R(X) denote the possible values of X and let R(Y) denote the possible values of To compute E[Y], we could irst find its probability mass function fy(v), and then compute ER(Y) But the Rule of the Lazy Statistician tells us that it is sufficient to compute Σ t(z) . Ix(z). TER(X) We will prove that this is the case a) Let () t)y. Write a formula for fy () using t b) Using your formula for fy (), prove that JER(Y) rER(X)

Answer 1

Let us begin with the discrete random variable X with probability mass function  $f_{X}(x)$ and here as given

$Y=toX=t(X)$

where the function is such that

$t : \mathbb{R}\rightarrow \mathbb{R}$

a) 1st part,

$t^{-1}(y)= \left \{ x:t(x)=y \right \}$

The probability mass function $f_{Y(y)}$ for  $Y=t(X)$

can be written in the form given below

$f_{Y(y)}=\sum_{x\epsilon t^{-1}(y)}^{}Rf_{X}(x)$

b) 2nd part,

$f_{Y(y)}=\sum_{x\epsilon t^{-1}(y)}^{}Rf_{X}(x)$

$\sum y.f_{Y}(y)=\sum t(x).f_{X}(x)$

where the range will change from $x\epsilon t^{-1}(y)$ to $x\epsilon R(X)$ and we know $t(x)=y$

$\sum y.f_{Y}(y)=\sum t(x).f_{X}(x)$ along with the range $x\epsilon R(X)$