How do we calculate the expected value of $Y$?

Let $\Omega = \{−1, 0, 1, 3\}$ with the probability function $p$ given by

I want to calculate the expected value and the variance of $X(\omega)=\omega$, $Y(\omega)=5\omega-3$, $Z(\omega)=(\omega-1)^2$.

For $X$ I have done the following :

We have that $X\in \{-1,0,1,3\}$.

The expected value is \begin{equation*}E[X]=\sum_{x\in X(\Omega)}xP[X=x]=(-1)\cdot \frac{1}{10}+0\cdot \frac{3}{10}+1\cdot \frac{2}{10}+3\cdot \frac{4}{10}= \frac{13}{10}\end{equation*} The variance is \begin{align*}Var[X]&=\sum_{x\in X(\Omega)}(x-E[X])P[X=x]\\ & =\left (-1-\frac{13}{10}\right )\cdot \frac{1}{10}+\left (0-\frac{13}{10}\right )\cdot \frac{3}{10}+\left (1-\frac{13}{10}\right )\cdot \frac{2}{10}+\left (3-\frac{13}{10}\right )\cdot \frac{4}{10}\\ & =0\end{align*}

Doe $Y$ we have $Y=5\omega-3\in \{-8,-3,2,12\}$, or not? But then we don't knowthe probabilities.

Or do weuse the property $E[aX+b]=aE[X]+b$ ?

Solution 1:

Your calculation of $E[X]$ is correct but the calculation of $\text{Var}[X]$ is not correct.

$ \text {Var}[X] = \sum(x - E[X])^2 \cdot P(X = x)$ which translates to,

$ \text {Var}[X] = E[X^2] - (E[X])^2 $

Now, $E[X^2] = (-1)^2 \cdot \dfrac{1}{10} + 1^2 \cdot \dfrac{2}{10} + 3^2 \cdot \dfrac{4}{10} = \dfrac{39}{10}$

So, $ \displaystyle \text {Var}[X] = \frac{39}{10} - \left(\frac{13}{10}\right)^2 = \frac{221}{100}$

Yes you can use the formula $~E[aX+b] = aE[X] + b$

Also note that $~ \text{Var} [aX+b] = a^2 \text{Var}[X]$

Solution 2:

You can use the formula $$\mathbb{E}[aX+b] = a\mathbb{E}[X]+b$$ to see that

$$\mathbb{E}[Y] = 5\cdot \frac{13}{10} -3 = \frac{7}{2}\ .$$

However, you can also calculate it directly because you do have the probabilities for $Y$. For example

$$\Pr(Y=-8) = \Pr(5\omega-3 = -8) = \Pr(\omega=-1)\ .$$

Since $Y$ is defined in terms of $\omega$, its distribution is also determined by the distribution for $\omega$. So once you get the distributions for $Y$ and $Z$ you can calculate their expectations and variances just like you did for $X$.