# Portfolio with several risky assets and one risk-free

Suppose we have $N$ risky assets $r_1$, $r_2$, ... , $r_N$ with a covariance matrix C. If we want to build a portfolio $\omega = (\omega_1, \ldots, \omega_N)^t$ (I loosely denote the portfolio with the weights vector $\omega$) with a minimum variance we solve the following optimization problem: $$\omega^tC\omega \longrightarrow \min$$ subject to $$\omega^t.1_N = 1 \\ \omega \geq 0_N$$ where $1_N$ and $0_N$ are column vectors of dimension $N$ with ones and zeros, respectively. This problem could be solved, for example, with Lagrange multipliers and the problem is a convex optimization problem.

Now let us add a risk-free asset $r_f$. If we now want to build a portfolio how does the optimization problem change? In my research I've seen that people refer to Sharpe ratio optimization but I couldn't identify the rigorous mathematical formulation, only strategies for a solution.

So how is the model above adapted?

• Mathematically it is the same formulation: minimizing $w^T C w$ subject to $w^T R=r_p$. But geometrically the picture changes: when P and Q are two risky assets in MV space, the combinations are on a CURVE connecting P and Q (parabola). When R is a risk-free asset and P is a risky asset, the combinations are on a STRAIGHT LINE connecting R and P. So the geometry oi the solution changes, algebraically it is the same problem, the same "model". – noob2 Apr 4 '17 at 13:30
• @noob2, what are $R$ and $r_p$ in your comment? – Veliko Apr 4 '17 at 13:34
• $R$ the expected return vector (of size N), $r_p$ the return on the portfolio (scalar). $C$ is the N by N covariance matrix. – noob2 Apr 4 '17 at 13:36
• @noob2 But we initially don't know $r_p$ and also how does the expected return $E(r_f)$ of the risk-free asset participate in that model? – Veliko Apr 4 '17 at 13:38
• The expected returns on the risky assets are $[r_1 r_2 \cdots r_N]=R$. When you add a riskless asset it can become the zero-th entry in this vector $[r_f r_1 \cdots r_N]$, I would call this $R'$ if I want to keep it separate from the other. In any case $r_f$ is just another expected return. – noob2 Apr 4 '17 at 14:14