Why does expectation of a transformation of one random variable appear different from expectation of one random variable?

Question

Let's say we have a continuous random variable $X_{1}$ with a probability density function $p_{X_{1}}$. The expectation is: $$E(X_1) = \int_{X_{1}} dx_{1} \Big ( x_{1} \cdot p_{X_{1}}(x_{1}) \Big)$$.

Now say we have a random variable $Y_{1}$, which is obtained as a transformation $g_{1}:X_{1} \rightarrow Y_{1}$ where $y_{1} = (x_{1} - 4)^{2} $. The expectation for $Y_{1}$ is: $$E(Y_{1})=\int_{X_{1}} dx_{1} \Big( (x_{1}-4)^2 \cdot p_{X_{1}}(x_{1}) \Big) $$.

Why is the expectation not: $$E(Y_{1})= \int_{Y_{1}} dy_{1} \Big( y_{1} \cdot p_{Y_{1}}(y_{1}) \Big) =\int_{X_{1}} dx_{1} \Big( (x_{1}-4)^2 \cdot p_{Y_{1}}((x_{1}-4)^2) \Big)$$? Does it mean $p_{Y_{1}}\big((x_{1}-4)^2\big) = p_{X_{1}}(x_{1}) $ and if so why is that the case?

Answer 1

Because $$E(Y_{1})= \int_{Y_{1}} dy_{1} \Big( y_{1} \cdot p_{Y_{1}}(y_{1}) \Big) \neq \int_{X_{1}} dx_{1} \Big( (x_{1}-4)^2 \cdot p_{Y_{1}}((x_{1}-4)^2) \Big)$$

you need to substitute the expression for $y_1$ into the differential too.