Expected values of squares

Question

Question A fair coin is tossed three times. Let Y be the random variable that denotes the square of the number of heads. For example, in the outcome HTH, there are two heads and Y = 4. What is E[Y]?

My answer:

possible outcomes to toss a coin three times : 0, 1, 2, 3 possible outcomes of Y : 0, 1, 4, 9 E[Y] = (1/6 * 0) + (1/6 * 1) + (1/6 * 4) + (1/6 * 9)

Is it ok? Thanks!

Answer 1

In general for $n$ tosses $$ \mathbf{E} = \sum_{i=0}^n{n \choose i}\left(\frac1{2}\right)^{\!\!n} i^2 $$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{h:\ HEAD.\ t:\ TOSSES}$.

\begin{align} &\bbox[10px,#ffd]{\sum_{h = 0}^{t}{t \choose h} \pars{1 \over 2}^{t - h}\pars{1 \over 2}^{h}h^{2}} = \left.{1 \over 2^{t}}\,\pars{x\,\partiald{}{x}}^{2}\sum_{h = 0}^{t} {t \choose h}x^{h}\,\right\vert_{\ x\ =\ 1} \\[5mm] = &\ \left.{1 \over 2^{t}}\,\pars{x\,\partiald{}{x}}^{2}\pars{1 + x}^{t}\,\right\vert_{\ x\ =\ 1} = \left.{1 \over 2^{t}}\,x\,\partiald{}{x}tx\pars{1 + x}^{t - 1} \,\right\vert_{\ x\ =\ 1} \\[5mm] = &\ {t \over 2^{t}}\,x\bracks{% \pars{1 + x}^{t - 1} + x\pars{t - 1}\pars{1 + x}^{t - 2}} _{\ x\ =\ 1} \\[5mm] = &\ {t \over 2^{t}}\bracks{2^{t - 1} + \pars{t - 1}2^{t - 2}} = {1 \over 2}\,t + {1 \over 4}\,t\pars{t - 1} = \bbx{t\pars{t + 1} \over 4} \end{align}

Answer 3

In this particular case the probabilities of 0, 1, 2, and 3 heads are 1/8, 3/8, 3/8, 1/8 and so

$$ E(Y) = (1/8) \times 0^2 + (3/8) \times 1^2 + (3/8) \times 2^2 + (1/8) \times 3^2 = 3. $$

In general for $n$ tosses, you have $E(Y) = E(X^2)$ where $X$ is a binomial(n, 1/2) random variable. Furthermore $E(X^2) = E(X)^2 + Var(X)$. It's well-known that $E(X) = n/2, Var(X) = n/4$, and so you have $E(X)^2 = (n/2)^2 + n/4 = n(n+1)/4$. Felix Marin already proved this by a different route.