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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.

User PLA
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The expected length of code for one encoded symbol is


\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha\ell_\alpha

where
p_\alpha is the probability of picking the letter
\alpha, and
\ell_\alpha is the length of code needed to encode
\alpha.
p_\alpha is given to us, and we have


\begin{cases}\ell_w=1\\\ell_x=2\\\ell_y=\ell_z=3\end{cases}

so that we expect a contribution of


\frac12+\frac24+\frac{2\cdot3}8=\frac{11}8=1.375

bits to the code per encoded letter. For a string of length
n, we would then expect
E[L]=1.375n.

By definition of variance, we have


\mathrm{Var}[L]=E\left[(L-E[L])^2\right]=E[L^2]-E[L]^2

For a string consisting of one letter, we have


\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha{\ell_\alpha}^2=\frac12+\frac{2^2}4+\frac{2\cdot3^2}8=\frac{15}4

so that the variance for the length such a string is


\frac{15}4-\left(\frac{11}8\right)^2=(119)/(64)\approx1.859

"squared" bits per encoded letter. For a string of length
n, we would get
\mathrm{Var}[L]=1.859n.

User Kiefer
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