Given that we know about random variable X and its probability density function and it's expectation is E(X). We know that when we calculate E(AX+B) we get AE(X)+B.
But when I did change of Variables, assumed Y=AX+B, and computed its probability density function it came out to be 1/A times the pdf of X. So According to me, when we calculate expectation of X i.e E(AX+B) we have to integrate Y*probability density of Y.. which gives the expectation as E(X)+B/Aenter image description here
This equation is wrong:
$$\mathfrak{f}_y(y)=\frac{d}{dy}(f_y(y))=\frac{d}{dy}\left(f_x\left(\frac{y-B}{A}\right)\right)=f_x(x)\cdot \frac{1}{A}$$
You accidentally turned $(y-B)/A$ into $x$ in the last step. Instead, it should be
$$\mathfrak{f}_y(y)=\frac{d}{dy}(f_y(y))=\frac{d}{dy}\left(f_x\left(\frac{y-B}{A}\right)\right)=\mathfrak{f}_x\left(\frac{y-B}{A}\right)\cdot \frac{1}{A}$$
Thus, we have
$$ \begin{align*} \mathbb{E}[Y] &= \int y\mathfrak{f}_y(y)dy \\ &= \int y \mathfrak{f}_x\left(\frac{y-B}{A}\right) \cdot \frac{1}{A}dy \\ &= \int (Ax+B) \mathfrak{f}_x(x)dx \\ &= A \int x\mathfrak{f}_x(x)dx + B\int \mathfrak{f}_x(x)dx \\ &= A\mathbb{E}[X] + B\cdot 1 \\ &= A\mathbb{E}[X]+B \end{align*} $$
as expected. Notice that the $1/A$ goes away when changing the integral from $dy$ to $dx$ because $x=(y-B)/A$, so $dx=(1/A)\cdot dy$.