Question

Since \(\displaystyle{P}\le{f}{t}{\left\lbrace{x}=-{1}{r}{i}{g}{h}{t}\right\rbrace}={P}\le{f}{t}{\left\lbrace{x}={1}{r}{i}{g}{h}{t}\right\rbrace}=\frac{{1}}{{2}}\) , calculate the expected value and variance of the X random variable.
\(\displaystyle{E}{\left[{\left({X}+{1}\right)}^{{{2}}}\right]}=\)?

Answer 1

Step 1
Solution:
From the given information,
\(\displaystyle{P}\le{f}{t}{\left\lbrace{x}=-{1}{r}{i}{g}{h}{t}\right\rbrace}={\frac{{{1}}}{{{2}}}}\)
\(\displaystyle{P}\le{f}{t}{\left\lbrace{x}={1}{r}{i}{g}{h}{t}\right\rbrace}={\frac{{{1}}}{{{2}}}}\)
Step 2
Then, the expected value of the X random variable is
\(\displaystyle{E}{\left({X}\right)}=\sum{x}{P}{\left({x}\right)}\)
\(\displaystyle={\left(-{1}\right)}\times{\frac{{{1}}}{{{2}}}}+{\left({1}\times{\frac{{{1}}}{{{2}}}}\right)}={0}\)
Thus, the expected value of the X random variable is 0.
Step 3
\(\displaystyle{E}{\left({X}^{{{2}}}\right)}=\sum{x}^{{{2}}}{P}{\left({x}\right)}\)
\(\displaystyle={\left({\left(-{1}\right)}^{{{2}}}\times{\frac{{{1}}}{{{2}}}}\right)}+{\left({1}^{{{2}}}\times{\frac{{{1}}}{{{2}}}}\right)}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}+{\frac{{{1}}}{{{2}}}}={1}\)
Then, the variance of the X random variable is
\(\displaystyle{V}{\left({X}\right)}={E}{\left({X}^{{{2}}}\right)}-\le{f}{t}{\left\lbrace{E}{\left({X}\right)}{r}{i}{g}{h}{t}\right\rbrace}^{{{2}}}\)
\(\displaystyle={1}-{0}^{{{2}}}={1}\)
Thus, the variance of the X random variable is 1.
Step 4
\(\displaystyle{E}{\left[{\left({X}+{1}\right)}^{{{2}}}\right]}={E}{\left[{X}^{{{2}}}+{2}{X}+{1}\right]}\)
\(\displaystyle={E}{\left[{X}^{{{2}}}\right]}+{2}{E}{\left[{X}\right]}+{E}{\left[{1}\right]}\)
\(\displaystyle={1}+{2}{\left({0}\right)}+{1}={2}\)