# Solved – the problem about Reverse Regression and how does the IV approach help to solve it

I really need some help with reverse regressions. I'm trying to solve the following exercise:

a) Consider the following Conditional Expectation Function model:
\$Y = alpha + beta X + epsilon , E[Y|X] = alpha + beta X\$

Suppose that we have transformed the above equation in the following way:
\$X = -(alpha /beta) + (1/beta)Y – (1/beta)epsilon \$

Show that this equation does not satisfy the definition of the CEF model.

b) Consider now the best linear predictor equation:
\$Y = alpha_L + beta_L X + epsilon_L\$

Consider as well the reverse best linear predictor equation:
\$X = alpha^*_L + beta^*_L Y + epsilon^*_L\$.

Show that in the Best Linear Predictor Setting \$beta_L ne 1/beta^*_L\$

c) Assume there is an Instrument Z. Show that \$beta^*_L = 1/beta^*_L\$ in the IV setting.

I would be really thankful if someone could explain me the idea behind the problem about reverse Regression and how it is related to the IV approach.

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I'm sorry: I don't understand what you mean by point c. I'll reply to points a and b.

a) \$X = -(alpha /beta) + (1/beta)Y – (1/beta)epsilon rightarrow E[X_i|Y_i]=-(alpha /beta) + (1/beta)Y_i\$, because \$X perp epsilon\$. To satisfy the CEF model, we need: \$E[- (1/beta)epsilon_i | Y_i]= 0 rightarrow – (1/beta)E[epsilon_i | Y_i] = 0 rightarrow (beta!=0, E[epsilon_i | Y_i] = 0)\$. Let's focus on the second condition: \$E[epsilon_i | Y_i]=0\$, and define \$W_i=epsilon_i*Y_i\$. By the Law of Total Expectation, we have: \$E[epsilon_i | Y_i]=0 rightarrow E[W_i]=E_{Y_i}[E[W_i|Y_i]]=E[Y_i*E[epsilon_i|Y_i]]=E[Y_i*0]=0rightarrow Cov(epsilon,Y)=E[epsilon*Y]-E[epsilon]E[Y]=0-0=0.\$ But, given \$Y = alpha + beta X + epsilon\$, then: \$Cov(epsilon,Y)=beta*Cov(X,epsilon)+Cov(epsilon,epsilon)=beta*0+Var(epsilon)=sigma^2>0\$. Thus, the condition \$E[epsilon_i | Y_i]=0\$ does not hold, so the CEF model is not satisfied. To be precise, the condition would only be satisfied in the degenerate case, i.e.: \$sigma^2=0rightarrow epsilon_i=0 forall i\$, i.e. in the deterministic setting.

b) Let's suppose \$beta_L=1/beta^*_L\$. Then, we get: \$Cov(X,Y)/Var(Y)=Var(X)/Cov(X,Y)rightarrow Cov(X,Y)^2=Var(X)*Var(Y)rightarrow [Corr(X,Y)*SD( X)*SD(Y)]^2=Corr(X,Y)^2*Var(X)*Var(Y)=Var(X)*Var(Y)rightarrow Corr(X,Y)=+1 bigvee Corr(X,Y)=-1.\$ Again, the condition only holds in case the relationship between \$X\$ and \$Y\$ is deterministic.

In fact, Galton (1886) showed that, if you normalize \$X\$ and \$Y\$ to have the same variance, then the coefficient in regressing \$Y\$ on \$X\$ or viceversa are the same, and equal to the coefficient of correlation (thus, between \$-1\$ and \$1\$). In case of perfect association and same variance of \$X\$ and \$Y\$, the coefficient would be \$+1 bigvee -1\$ indeed. This link may help: http://davegiles.blogspot.it/2014/11/reverse-regression-follow-up.html

Also, you may be interested in these Qs&As:

Galton, Francis (1886): “Regression Towards Mediocrity in Hereditary Stature,” The Journal of the Anthropological Institute of Great Britain and Ireland, 15, 246-263.

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# Solved – the problem about Reverse Regression and how does the IV approach help to solve it

I really need some help with reverse regressions. I'm trying to solve the following exercise:

a) Consider the following Conditional Expectation Function model:
\$Y = alpha + beta X + epsilon , E[Y|X] = alpha + beta X\$

Suppose that we have transformed the above equation in the following way:
\$X = -(alpha /beta) + (1/beta)Y – (1/beta)epsilon \$

Show that this equation does not satisfy the definition of the CEF model.

b) Consider now the best linear predictor equation:
\$Y = alpha_L + beta_L X + epsilon_L\$

Consider as well the reverse best linear predictor equation:
\$X = alpha^*_L + beta^*_L Y + epsilon^*_L\$.

Show that in the Best Linear Predictor Setting \$beta_L ne 1/beta^*_L\$

c) Assume there is an Instrument Z. Show that \$beta^*_L = 1/beta^*_L\$ in the IV setting.

I would be really thankful if someone could explain me the idea behind the problem about reverse Regression and how it is related to the IV approach.

I'm sorry: I don't understand what you mean by point c. I'll reply to points a and b.

a) \$X = -(alpha /beta) + (1/beta)Y – (1/beta)epsilon rightarrow E[X_i|Y_i]=-(alpha /beta) + (1/beta)Y_i\$, because \$X perp epsilon\$. To satisfy the CEF model, we need: \$E[- (1/beta)epsilon_i | Y_i]= 0 rightarrow – (1/beta)E[epsilon_i | Y_i] = 0 rightarrow (beta!=0, E[epsilon_i | Y_i] = 0)\$. Let's focus on the second condition: \$E[epsilon_i | Y_i]=0\$, and define \$W_i=epsilon_i*Y_i\$. By the Law of Total Expectation, we have: \$E[epsilon_i | Y_i]=0 rightarrow E[W_i]=E_{Y_i}[E[W_i|Y_i]]=E[Y_i*E[epsilon_i|Y_i]]=E[Y_i*0]=0rightarrow Cov(epsilon,Y)=E[epsilon*Y]-E[epsilon]E[Y]=0-0=0.\$ But, given \$Y = alpha + beta X + epsilon\$, then: \$Cov(epsilon,Y)=beta*Cov(X,epsilon)+Cov(epsilon,epsilon)=beta*0+Var(epsilon)=sigma^2>0\$. Thus, the condition \$E[epsilon_i | Y_i]=0\$ does not hold, so the CEF model is not satisfied. To be precise, the condition would only be satisfied in the degenerate case, i.e.: \$sigma^2=0rightarrow epsilon_i=0 forall i\$, i.e. in the deterministic setting.

b) Let's suppose \$beta_L=1/beta^*_L\$. Then, we get: \$Cov(X,Y)/Var(Y)=Var(X)/Cov(X,Y)rightarrow Cov(X,Y)^2=Var(X)*Var(Y)rightarrow [Corr(X,Y)*SD( X)*SD(Y)]^2=Corr(X,Y)^2*Var(X)*Var(Y)=Var(X)*Var(Y)rightarrow Corr(X,Y)=+1 bigvee Corr(X,Y)=-1.\$ Again, the condition only holds in case the relationship between \$X\$ and \$Y\$ is deterministic.

In fact, Galton (1886) showed that, if you normalize \$X\$ and \$Y\$ to have the same variance, then the coefficient in regressing \$Y\$ on \$X\$ or viceversa are the same, and equal to the coefficient of correlation (thus, between \$-1\$ and \$1\$). In case of perfect association and same variance of \$X\$ and \$Y\$, the coefficient would be \$+1 bigvee -1\$ indeed. This link may help: http://davegiles.blogspot.it/2014/11/reverse-regression-follow-up.html

Also, you may be interested in these Qs&As:

Galton, Francis (1886): “Regression Towards Mediocrity in Hereditary Stature,” The Journal of the Anthropological Institute of Great Britain and Ireland, 15, 246-263.

Rate this post