The top right comers of the highest rectangle (modal class rectangle) and the preceding rectangle are joined by a straight line. Similarly, the top left corners of the-highest rectangle and the rectangle just on its right are joined by a straight line. The value of the mode is given by the point of intersection of the two lines graphically.

Please refer to link.

The equation of the mode is derived by making use of the fact $\triangle AEB\sim \triangle DEC$ and $\triangle BEF \sim \triangle BDC$ the formula for the mode is derived $$ Mode=l_o+x=l_o+\frac{f_1-f_o}{(f_1-f_o)+(f_1-f_2)}.h $$

Why do we get the value of the mode as the intersection of the lines obtained by joining the corners of the rectangles of the modal class and the neighbouring classes ? What assumptioms are taken here to obtain the formula ?


In truth, this is only one of many possible estimates for the mode, when the data are binned/grouped. You could construct a continuous probability distribution, and depending on how you discretize or "bin" the outcomes, you could get different modes.

Let us illustrate with an example. Suppose $$X \sim \operatorname{Gamma}(3, 1)$$ with density $$f_{X}(x) = \frac{x^2}{2} e^{-x}, \quad x > 0.$$ The true mode of this distribution is found by computing the derivative and looking for critical points: $$0 = f'(x) = -\frac{x^2}{2}(x-2) e^{-x},$$ hence $x = 2$ is the exact mode.

Now suppose we discretize the density into integer width bins, i.e., let $$Y = \lfloor X \rfloor,$$ so that $$\Pr[Y = y] = \Pr[y \le X < y+1] = \frac{1}{2} \int_{x=y}^{y+1} x^2 e^{-x} \, dx.$$ This is not difficult to compute exactly: $$\Pr[Y = y] = \frac{e^{-(y+1)}}{2} \left(-5 + 2e + 2(e-2)y + (e-1)y^2\right).$$ From this, we can compute $$\Pr[Y = 1] = \frac{5(e-2)}{2e^2} = f_0, \\ \Pr[Y = 2] = \frac{10e-17}{2e^3} = f_1, \\ \Pr[Y = 3] = \frac{17e-26}{2e^4} = f_2.$$ Using the formula provided, and with $l_0 = 2$, we have compute the mode as $$2 + \frac{\frac{10e-17}{2e^3} - \frac{5(e-2)}{2e^2}}{\frac{10e-17}{2e^3} - \frac{5(e-2)}{2e^2} + \frac{10e-17}{2e^3} - \frac{17e-26}{2e^4}} \cdot \frac{10e-17}{2e^3} \approx 2.0336342,$$ but we already knew that this calculation would yield a number strictly greater than $2$.

If, however, we binned the data differently, e.g. $$W = \lfloor X + 1/2 \rfloor,$$ we have $$\Pr[W = w] = \Pr[w - 1/2 \le X < w + 1/2] = \frac{e^{-(w+1/2)}}{8} \left(-13 + 5e + 4(e-3)w + 4(e-1) w^2\right),$$ and the resulting estimate for the mode is (calculations omitted) $1.67949$.

So what we can take away from this is that when data are binned from an underlying continuous distribution, you really can't tell where the mode is within the bin with the highest count, or even if the bin with the highest count actually contains the true mode.

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  • $\begingroup$ could you please comment on the assumptions taken to obtain the above equations for mode ? $\endgroup$ – ss1729 May 22 at 23:38
  • $\begingroup$ @ss1729 I cannot, because I do not see a rationale for the estimation method used. As I have tried to illustrate, this method yields an estimate that may not reflect the true mode because in general, any binning of a continuous distribution will remove information about the true mode. There are other possible estimators, but none can recover what is lost. $\endgroup$ – heropup May 22 at 23:41
  • $\begingroup$ i think it might be misleading since I used the term "continuous frequency". I actually ment to say "grouped data". A similar post is asked about it but does not address my doubt, check math.stackexchange.com/questions/905347/… $\endgroup$ – ss1729 May 22 at 23:45
  • $\begingroup$ @ss1729 I did not misunderstand you. Grouped data = binned data. The act of grouping data destroys information about the mode, irrespective of whether the underlying distribution is continuous or discrete or some mix of the two. What you seem to be not understanding despite my repeated explanations is that this is estimator does not have any underlying rationale because as I have pointed out, the mode does not even need to be contained in the bin with the highest observed frequency. $\endgroup$ – heropup May 22 at 23:56
  • $\begingroup$ okay, so what about the above formula ? is it completely useless ? $\endgroup$ – ss1729 May 23 at 0:24

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