What is Mean Square Error?

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The mean square error (MSE) is defined as mean or average of the square of the difference between actual and estimated values.

Mathematically, the mean square error is,

$$\mathrm{\varepsilon =\frac{1}{t_{2}-t_{1}}\int_{t_{1}}^{t_{2}}\left [ x(t) -\sum_{r=1}^{n}C_{r}g_{r}(t)\right ]^{2}dt}$$

$$\mathrm{\varepsilon =\frac{1}{t_{2}-t_{1}}\left [ \int_{t_{1}}^{t_{2}}x^{2}(t)dt+\sum_{r=1}^{n}C_{r}^{2}\int_{t_{1}}^{t_{2}}g_{r}^{2}(t)dt-2\sum_{r=1}^{n}C_{r}\int_{t_{1}}^{t_{2}}x(t)g_{r}(t)dt\right ]\; ...(1)}$$

$$\mathrm{\therefore C_{r}=\frac{\int_{t_{1}}^{t_{2}}x(t)g_{r}(t)dt}{\int_{t_{1}}^{t_{2}}g_{r}^{2}(t)dt}=\frac{1}{K_{r}}\int_{t_{1}}^{t_{2}}x(t)g_{r}(t)dt\; \; ...(2)}$$

$$\mathrm{\therefore \int_{t_{1}}^{t_{2}}x(t)g_{r}(t)dt=C_{r}\int_{t_{1}}^{t_{2}}g_{r}^{2}(t)dt=C_{r}K_{r}\; \; ...(3)}$$

Using equations (1) and (3), we have,

$$\mathrm{\varepsilon =\frac{1}{t_{2}-t_{1}}\left [\int_{t_{1}}^{t_{2}} x^{2}(t)dt +\sum_{r=1}^{n}C^{2}_{r}K_{r}-2\sum_{r=1}^{n}C^{2}_{r}K_{r}\right ]}$$

$$\mathrm{\Rightarrow \varepsilon =\frac{1}{t_{2}-t_{1}}\left [\int_{t_{1}}^{t_{2}} x^{2}(t)dt -\sum_{r=1}^{n}C^{2}_{r}K_{r}\right ]\; \; ...(4)}$$

$$\mathrm{\Rightarrow \varepsilon =\frac{1}{t_{2}-t_{1}}\left [ \int_{t_{1}}^{t_{2}}x^{2}(t)dt-(C_{1}^{2}K_{1}+C_{2}^{2}K_{2}+\cdot \cdot \cdot +C_{n}^{2}K_{n}) \right ]\; \; \cdot \cdot \cdot (5)}$$

Therefore, the mean square error can be evaluated using eqn. (5).

Numerical Example

A rectangular function is defined as,

$$\mathrm{x(t)=\left\{\begin{matrix} 1\; \; for\, 0< t< \Pi \ -1\; \; for\, \Pi< t< 2\Pi\ \end{matrix}\right.}$$

The signal x(t) is approximated to a sinusoidal function $\mathrm{x(t)=\frac{4}{\Pi }\sin t}$ sin 𝑡 over the interval [0, 2π]. Evaluate the mean square error in this approximation.

Solution

The approximation of the rectangular function x(t) by a sinusoidal signal sin 𝑡 is shown in the figure and is given by,

$$\mathrm{x(t)=\frac{4}{\Pi }\sin t}$$

The mean square error in this approximation can be evaluated using the formula,

$$\mathrm{\Rightarrow \varepsilon =\frac{1}{t_{2}-t_{1}}\left [\int_{t_{1}}^{t_{2}} x^{2}(t)\: dt-\int_{t_{1}}^{t_{2}} \left ( \frac{4}{\pi }\sin t \right )^{2}\; dt\right] }$$

Here, 𝑡1 = 0 and 𝑡2 = 2𝜋, therefore,

$$\mathrm{\Rightarrow \varepsilon =\frac{1}{2\pi -0}\left [\int_{0}^{2\pi } 1\: dt-(\frac{4}{\pi })^{2}\int_{0}^{2\pi } \sin^{2} t\; dt\right] }$$

$$\mathrm{\Rightarrow \varepsilon =\frac{1}{2\pi}\left [\int_{0}^{2\pi }\: dt-\left ( \frac{4}{\pi } \right )^{2}\int_{0}^{2\pi } \left ( \frac{1-cos2t}{2} \right )\; dt\right] }$$

$$\mathrm{\Rightarrow \varepsilon =\frac{1}{2\pi}\left [\int_{0}^{2\pi }\: dt-\left ( \frac{16}{\pi^{2} } \right )\int_{0}^{2\pi } \left ( \frac{1-cos2t}{2} \right )\; dt\right] }$$

$$\mathrm{\Rightarrow \varepsilon =\frac{1}{2\pi}\left [\left [ t \right ]_{0}^{2\pi }-\frac{16}{2\pi ^{2}}\left [ t-\frac{sin2t}{2} \right ]_{0}^{2\pi }\right] }$$

$$\mathrm{\Rightarrow \varepsilon =\frac{1}{2\pi }\left [ \left ( 2\pi -0 \right )-\frac{8}{\pi ^{2}}\begin{Bmatrix} (2\pi -0)-\left ( \frac{sin4\pi -sin0}{2} \right )\ \end{Bmatrix} \right ]}$$

$$\mathrm{\Rightarrow \varepsilon =\frac{1}{2\pi }\left [ 2\pi -\frac{16}{\pi} \right ]=1-\frac{8}{\pi ^{2}}=0.189}$$

∴ Mean square error, 𝜀 = 0.189 = 18.9%

Updated on 13-Nov-2021 10:36:55