Expectation of Linear Transformation of Random Variable/Continuous
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Theorem
Let $X$ be a continuous random variable.
Let $a, b$ be real numbers.
Then we have:
- $\expect {a X + b} = a \expect X + b$
where $\expect X$ denotes the expectation of $X$.
Proof
Let $\map \supp X$ be the support of $X$.
Let $f_X : \map \supp X \to \R$ be the probability density function of $X$.
The validity of the material on this page is questionable. In particular: What if the density $f_X$ does not exist? You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Questionable}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Then:
\(\ds \expect {a X + b}\) | \(=\) | \(\ds \int_{x \mathop \in \map \supp X} \paren {a x + b} \map {f_X} x \rd x\) | Expectation of Function of Continuous Random Variable | |||||||||||
\(\ds \) | \(=\) | \(\ds a \int_{x \mathop \in \map \supp X} x \map {f_X} x \rd x + b \int_{x \mathop \in \map \supp X} \map {f_X} x \rd x\) | Linear Combination of Definite Integrals | |||||||||||
\(\ds \) | \(=\) | \(\ds a \expect X + b \times 1\) | Definition of Expectation of Continuous Random Variable | |||||||||||
\(\ds \) | \(=\) | \(\ds a \expect X + b\) |
$\blacksquare$