Critiques of the Two-Sector Mahalanobis Model

Abstract

The Mahalanobis model attracted worldwide attention by academic economists, statisticians, and development practitioners in the field. In this chapter, we detail what we feel are the most important of these critiques, due to Martin Bronfenbrenner, Charles Bettelheim, Hannan Ezikiel, and J.B.S. Haldane. We discuss the implications of these criticisms for the Mahalanobis model both analytically and computationally. These critiques are based on the Mahalanobis model outlined in Chap. 2, which is a closed-economy model.

4.7 Technical Appendix

4.A Mahalanobis Model v/s Bettelheim Model—Comparison of Output for a given $\lambda _{i}$ : From Eqs. (2.20) and (4.26), suppose $Y_{t}^{Bet}$ and $Y_{t}^{MM}$ denote the output in the Bettelheim model and the Mahalanobis model, respectively.²³

$$\begin{aligned} Y_{t}^{Bet}=Y_{0}\left[ 1+\alpha _{0}\left( \lambda _{i}\beta _{i}+\lambda _{c}\beta _{c}\right) \left( \frac{\left( 1+\lambda _{i}\beta _{i}-\frac{\lambda _{c}}{mc}\right) ^{t}-1}{\lambda _{i}\beta _{i}-\frac{\lambda _{c}}{mc}}\right) \right] \end{aligned}$$

$$\begin{aligned} Y_{t}^{MM}=Y_{0}\left[ 1+\alpha _{0}\left( \lambda _{i}\beta _{i}+\lambda _{c}\beta _{c}\right) \left( \frac{\left( 1+\lambda _{i}\beta _{i}\right) ^{t}-1}{\lambda _{i}\beta _{i}}\right) \right] . \end{aligned}$$

For a given $\lambda _{i}$ in an economy, let us compare the absolute value of net outputs in the two models.

If $\frac{1}{mc}=0$, then Mahalanobis model gives the same national income as the Bettelheim model, i.e., $Y_{t}^{MM}=Y_{t}^{Bet}$ $\forall t.$ Also, $Y_{t}^{MM}=Y_{t}^{Bet}$ $\forall mc$ for $t=1.$

For $t>1,$ intuitively, both numerator and denominator of $Y_{t}^{Bet}$ are less than that of $Y_{t}^{MM}.$ But since the numerator is raised to the power t,  the negative impact of lower numerator is higher than the positive effect of lower denominator. This makes $Y_{t}^{Bet}<Y_{t}^{MM}.$ Suppose $a = \alpha _0$, $x = \lambda _i\beta _i$, and suppose,

$Z=a\left( \frac{\left( 1+x\right) ^{t}-1}{x}\right) .$ Differentiating Z with respect to x, 

$$\begin{aligned} \frac{dZ}{dx}=\frac{a}{x^{2}}\left[ ((t-1)x-1)(1+x)^{t-1}+1\right] >0. \end{aligned}$$

Since, Z is increasing in x,  this implies

$$\begin{aligned} \alpha _{0}\left( \lambda _{i}\beta _{i}+\lambda _{c}\beta _{c}\right) \left( \frac{\left( 1+\lambda _{i}\beta _{i}-\frac{\lambda _{c}}{mc}\right) ^{t}-1}{\lambda _{i}\beta _{i}-\frac{\lambda _{c}}{mc}}\right) <\alpha _{0}\left( \lambda _{i}\beta _{i}+\lambda _{c}\beta _{c}\right) \left( \frac{\left( 1+\lambda _{i}\beta _{i}\right) ^{t}-1}{\lambda _{i}\beta _{i}}\right) , \end{aligned}$$

i.e., $Y_{t}^{MM}>Y_{t}^{Bet}$ for a given $\lambda _{i}$ in an economy for all values of mc and $t>1.$

Now, comparing the relative net outputs in the two models for a given $\lambda _{i}$,

$$\begin{aligned} \left( \frac{Y_{t+1}}{Y_{t}}\right) ^{MM}=\frac{\lambda _{i}\beta _{i}+\alpha _{0}\beta \left[ \left( 1+\lambda _{i}\beta _{i}\right) ^{t+1}-1\right] }{\lambda _{i}\beta _{i}+\alpha _{0}\beta \left[ \left( 1+\lambda _{i}\beta _{i}\right) ^{t}-1\right] } \end{aligned}$$

$$\begin{aligned} \left( \frac{Y_{t+1}}{Y_{t}}\right) ^{Bet}=\frac{\lambda _{i}\beta _{i}-\frac{\lambda _{c}}{mc}+\alpha _{0}\beta \left[ \left( 1+\lambda _{i}\beta _{i}-\frac{\lambda _{c}}{mc}\right) ^{t+1}-1\right] }{\lambda _{i}\beta _{i}-\frac{\lambda _{c}}{mc}+\alpha _{0}\beta \left[ \left( 1+\lambda _{i}\beta _{i}-\frac{\lambda _{c}}{mc}\right) ^{t}-1\right] }. \end{aligned}$$

We get $\left( \frac{Y_{t+1}}{Y_{t}}\right) ^{MM}=\left( \frac{Y_{t+1}}{Y_{t}}\right) ^{Bet}$ if $\frac{1}{mc}=0$ and $\left( \frac{Y_{t+1}}{Y_{t}}\right) ^{MM}>\left( \frac{Y_{t+1}}{Y_{t}}\right) ^{Bet}$ if $\frac{1}{mc}>0$ because

$$\begin{aligned} \frac{\lambda _{i}\beta _{i}+\alpha _{0}\beta \left[ \left( 1+\lambda _{i}\beta _{i}\right) ^{t+1}-1\right] }{\lambda _{i}\beta _{i}+\alpha _{0}\beta \left[ \left( 1+\lambda _{i}\beta _{i}\right) ^{t}-1\right] }\gtreqless \frac{\lambda _{i}\beta _{i}-\frac{\lambda _{c}}{mc}+\alpha _{0}\beta \left[ \left( 1+\lambda _{i}\beta _{i}-\frac{\lambda _{c}}{mc}\right) ^{t+1}-1\right] }{\lambda _{i}\beta _{i}-\frac{\lambda _{c}}{mc}+\alpha _{0}\beta \left[ \left( 1+\lambda _{i}\beta _{i}-\frac{\lambda _{c}}{mc}\right) ^{t}-1\right] }. \end{aligned}$$

By cross multiplication and canceling the terms, we get

$$\begin{aligned} \left( 1+\lambda _{i}\beta _{i}\right) ^{t}\left[ \begin{array}{c} \left( \lambda _{i}\beta _{i}\right) ^{2}-\frac{\lambda _{i}\beta _{i}\lambda _{c}}{mc}- \\ \alpha _{0}\lambda _{i}\beta _{i}\beta +\frac{\alpha _{0}\lambda _{c}\beta }{mc}\left( 1+\lambda _{i}\beta _{i}-\frac{\lambda _{c}}{mc}\right) ^{t}\end{array}\right] \gtreqless \left( 1+\lambda _{i}\beta _{i}-\frac{\lambda _{c}}{mc}\right) ^{t}\left[ \begin{array}{c} \left( \lambda _{i}\beta _{i}\right) ^{2}-\frac{\lambda _{i}\beta _{i}\lambda _{c}}{mc} \\ -\alpha _{0}\lambda _{i}\beta _{i}\beta +\frac{\alpha _{0}\lambda _{c}\beta }{mc}\end{array}\right] . \end{aligned}$$

Therefore, as long as $\lambda _{i}\beta _{i}-\frac{\lambda _{c}}{mc}>0,$ $\left( 1+\lambda _{i}\beta _{i}-\frac{\lambda _{c}}{mc}\right) ^{t}>1$ and $\left( 1+\lambda _{i}\beta _{i}\right) ^{t}>\left( 1+\lambda _{i}\beta _{i}-\frac{\lambda _{c}}{mc}\right) ^{t},$ $L.H.S>R.H.S.$ This implies that $\left( \frac{Y_{t+1}}{Y_{t}}\right) ^{MM}>\left( \frac{Y_{t+1}}{Y_{t}}\right) ^{Bet}$ $\forall t$ and $\forall \frac{1}{mc}>0$ for a given $\lambda _{i}$ in an economy.

4.B Mahalanobis Model versus Bettelheim Model—Comparison of Output at respective critical values: At $\lambda _{i}=\frac{\alpha _{0}\beta _{c}}{(1-\alpha _{0})\beta _{i}+\alpha _{0}\beta _{c}}=\widehat{\lambda }_{i},$

$$\begin{aligned} Y_{t}^{MM}=Y_{0}\left[ 1+\alpha _{0}\beta ^{MM}\left( \frac{\left( 1+\widehat{\lambda }_{i}\beta _{i}\right) ^{t}-1}{\widehat{\lambda }_{i}\beta _{i}}\right) \right] , \end{aligned}$$

where $\beta ^{MM}=\widehat{\lambda }_{i}\beta _{i}+(1-\widehat{\lambda }_{i})\beta _{c}.$ Substituting the value of $\widehat{\lambda }_{i},$ we get

$$\begin{aligned} \beta ^{MM}=\frac{\alpha _{0}\beta _{c}\beta _{i}}{(1-\alpha _{0})\beta _{i}+\alpha _{0}\beta _{c}}+\frac{(1-\alpha _{0})\beta _{i}\beta _{c}}{(1-\alpha _{0})\beta _{i}+\alpha _{0}\beta _{c}}=\frac{\beta _{i}\beta _{c}}{(1-\alpha _{0})\beta _{i}+\alpha _{0}\beta _{c}} \end{aligned}$$

and

$$\begin{aligned} \widehat{\lambda }_{i}\beta _{i}=\frac{\alpha _{0}\beta _{c}\beta _{i}}{(1-\alpha _{0})\beta _{i}+\alpha _{0}\beta _{c}}. \end{aligned}$$

Therefore, we get

$$\begin{aligned} Y_{t}^{MM}=Y_{0}\left( 1+\widehat{\lambda }_{i}\beta _{i}\right) ^{t}. \end{aligned}$$

At $\lambda _{i}=\frac{\frac{1}{mc}+\alpha _{0}\beta _{c}}{\frac{1}{mc}+(1-\alpha _{0})\beta _{i}+\alpha _{0}\beta _{c}}=\lambda _{i}^{*},$

$$\begin{aligned} Y_{t}^{Bet}=Y_{0}\left[ 1+\alpha _{0}\beta ^{Bet}\left( \frac{\left( 1+\lambda _{i}^{*}\beta _{i}-\frac{\lambda _{c}^{*}}{mc}\right) ^{t}-1}{\lambda _{i}^{*}\beta _{i}-\frac{\lambda _{c}^{*}}{mc}}\right) \right] , \end{aligned}$$

where $\beta ^{Bet}=\lambda _{i}^{*}\beta _{i}+(1-\lambda _{i}^{*})\beta _{c}.$ Substituting the value of $\lambda _{i}^{*},$ we get

$$\begin{aligned} \beta ^{Bet}=\frac{\frac{\beta _{i}}{mc}+\alpha _{0}\beta _{c}\beta _{i}}{\frac{1}{mc}+(1-\alpha _{0})\beta _{i}+\alpha _{0}\beta _{c}}+\frac{(1-\alpha _{0})\beta _{i}\beta _{c}}{\frac{1}{mc}+(1-\alpha _{0})\beta _{i}+\alpha _{0}\beta _{c}}=\frac{\frac{\beta _{i}}{mc}+\beta _{c}\beta _{i}}{\frac{1}{mc}+(1-\alpha _{0})\beta _{i}+\alpha _{0}\beta _{c}} \end{aligned}$$

and

$$\begin{aligned} \lambda _{i}^{*}\beta _{i}-\frac{\lambda _{c}^{*}}{mc}=\frac{\frac{\beta _{i}}{mc}+\alpha _{0}\beta _{c}\beta _{i}}{\frac{1}{mc}+(1-\alpha _{0})\beta _{i}+\alpha _{0}\beta _{c}}-\frac{\frac{(1-\alpha _{0})\beta _{i}}{mc}}{\frac{1}{mc}+(1-\alpha _{0})\beta _{i}+\alpha _{0}\beta _{c}}=\frac{\frac{\alpha _{0}\beta _{i}}{mc}+\alpha _{0}\beta _{c}\beta _{i}}{\frac{1}{mc}+(1-\alpha _{0})\beta _{i}+\alpha _{0}\beta _{c}}. \end{aligned}$$

Therefore, we get

$$\begin{aligned} Y_{t}^{Bet}=Y_{0}\left( 1+\lambda _{i}^{*}\beta _{i}-\frac{\lambda _{c}^{*}}{mc}\right) ^{t}, \end{aligned}$$

where $\lambda _{c}^{*}=1-\lambda _{i}^{*}\$is the allocation of investment goods towards production of the consumer goods analogous to the critical value $\lambda _{i}^{*}.$ Now, both numerator and denominator of $\lambda _{i}^{*}\beta _{i}-\frac{\lambda _{c}^{*}}{mc}$ are greater than that of $\widehat{\lambda }_{i}\beta _{i}$ but numerator is greater by a smaller amount, and hence $\lambda _{i}^{*}\beta _{i}-\frac{\lambda _{c}^{*}}{mc}<\widehat{\lambda }_{i}\beta _{i}.$ This implies that at their respective critical values, $Y_{t}^{MM}>Y_{t}^{Bet}$ $\forall t$ and $\forall mc.$

4.C Solving Haldane’s Model: In Eq. (4.42), suppose $bxt=y$

$$\begin{aligned} ae^{-y}= & {} a-y[a-(a-b)x] \\ e^{-y}= & {} 1-y\left[ 1-\left( \frac{a-b}{a}\right) x\right] \\ \left( \frac{a-b}{a}\right) x= & {} 1-\left( 1-e^{-y}\right) y^{-1} \\= & {} 1-\left( 1-\left[ 1-y+\frac{y^{2}}{2!}-\frac{y^{3}}{3!}+.....\right] \right) y^{-1} \\= & {} 1-\left( y-\frac{y^{2}}{2!}+\frac{y^{3}}{3!}-\frac{y^{4}}{4!}+.....\right) y^{-1} \\= & {} 1-\left( 1-\frac{y}{2!}+\frac{y^{2}}{3!}-\frac{y^{3}}{4!}+....\right) \\= & {} \frac{y}{2!}-\frac{y^{2}}{3!}+\frac{y^{3}}{4!}-\frac{y^{4}}{5!}+.... \end{aligned}$$

Suppose $\frac{2(a-b)}{a}x=z$. We get Eq. (4.43).

$$\begin{aligned} z=2\left( \frac{y}{2!}-\frac{y^{2}}{3!}+\frac{y^{3}}{4!}-\frac{y^{4}}{5!}+....\right) . \end{aligned}$$

Now, to invert the series so that we can represent y in terms of z such as in Eq. (4.44), we can follow the following procedure:

Re-write (4.43) as

$$\begin{aligned} z=a_{1}y+a_{2}y^{2}+a_{3}y^{3}+..... \end{aligned}$$

(4.47)

Let the following be the corresponding inverted series:

$$\begin{aligned} y=A_{1}z+A_{2}z^{2}+A_{3}z^{3}+..... \end{aligned}$$

(4.48)

Substituting (4.48) in (4.47),

$$\begin{aligned} z= & {} a_{1}(A_{1}z+A_{2}z^{2}+A_{3}z^{3}+.....)+a_{2}(A_{1}z+A_{2}z^{2}+A_{3}z^{3}+.....)^{2} \\&+a_{3}(A_{1}z+A_{2}z^{2}+A_{3}z^{3}+.....)^{3}+.... \\ z= & {} (a_{1}A_{1})z+(a_{1}A_{2}+a_{2}A_{1}^{2})z^{2}+(a_{1}A_{3}+a_{3}A_{1}^{3}+2a_{2}A_{1}A_{2})z^{3}+....... \end{aligned}$$

Comparing parameters on both sides of the equation, we get

$$\begin{aligned} a_{1}A_{1}= & {} 1\Rightarrow A_{1}=\frac{1}{a_{1}} \\ a_{1}A_{2}+a_{2}A_{1}^{2}= & {} 0\Rightarrow A_{2}=-\frac{a_{2}}{a_{1}^{3}} \\ a_{1}A_{3}+a_{3}A_{1}^{3}+2a_{2}A_{1}A_{2}= & {} 0\Rightarrow A_{3}=\frac{2a_{2}^{2}-a_{1}a_{3}}{a_{1}^{5}} \end{aligned}$$

and so on.

From (4.43), $a_{1}=1,$ $a_{2}=-\frac{1}{3},$ $a_{3}=\frac{1}{12}$ and so on.

Substituting these values in the above-derived equations of $A_{1},A_{2},A_{3},....$, we get

$$\begin{aligned} A_{1}=1,A_{2}=\frac{1}{3},A_{3}=\frac{5}{36},\text { and so on} \end{aligned}$$

Substituting these values back in Eq. (4.48), we get the inverted series in Eq. (4.44):

$$\begin{aligned} y=z+\frac{1}{3}z^{2}+\frac{5}{36}z^{3}+..... \end{aligned}$$