A source of information randomly generates symbols from a four letter alphabet {w, x, y, z }. The probability of each symbol is as follows: P(w) = 1 2 ; P(x) = 1 4 ; P(y) = 1 8 ; and P(z) = 1 8 . These symbols are now encoded into binary codes using the scheme shown below. Let the random variable L denote the length of the binary code and pL(l) denote the PMF of L.symbol codew 0x 10y 110z 111Find the expectation and variance of L.

The expected length of code for one encoded symbol is[tex]\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha\ell_\alpha[/tex]where [tex]p_\alpha[/tex] is the probability of picking the letter [tex]\alpha[/tex], and [tex]\ell_\alpha[/tex] is the length of code needed to encode [tex]\alpha[/tex]. [tex]p_\alpha[/tex] is given to us, and we have[tex]\begin{cases}\ell_w=1\\\ell_x=2\\\ell_y=\ell_z=3\end{cases}[/tex]so that we expect a contribution of[tex]\dfrac12+\dfrac24+\dfrac{2\cdot3}8=\dfrac{11}8=1.375[/tex]bits to the code per encoded letter. For a string of length [tex]n[/tex], we would then expect [tex]E[L]=1.375n[/tex].By definition of variance, we have[tex]\mathrm{Var}[L]=E\left[(L-E[L])^2\right]=E[L^2]-E[L]^2[/tex]For a string consisting of one letter, we have[tex]\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha{\ell_\alpha}^2=\dfrac12+\dfrac{2^2}4+\dfrac{2\cdot3^2}8=\dfrac{15}4[/tex]so that the variance for the length such a string is[tex]\dfrac{15}4-\left(\dfrac{11}8\right)^2=\dfrac{119}{64}\approx1.859[/tex]"squared" bits per encoded letter. For a string of length [tex]n[/tex], we would get [tex]\mathrm{Var}[L]=1.859n[/tex].