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(a) If $m_k \sim p(m_k)$, where $p(m_k)$ is depicted in  the figure, and $m_k=n+h$, figure out $p_n(n)$ and   $p_h(h)$. \\\\  Ans:\\ ANS:\\  Becasue $m_k=n+h$ and $p_z(z)=p_x(x)*p_y(y)$, I could derive that $p(m_k)=p(n+h)=p_n(n)*p_h(h)$.  $p(m_k)$ is a symmetrical triangular-shaped distribution, and, it is the result of convolution of   two identical uniform distribution. Therefor, $p_n(n)$ and $p_h(h)$ can be two identical uniform distribution as: