Mathematics Homework

Assignment /Homework
We have two equations:

We need to find the relationship (positive or negative) between (X and n) as a full algebraic proof, where 0 σ and n  , 1 A , , (0,1) α β , and 0 1 δ   .
I’ve already found that the relationship between Y and n is negative by taking the derivative, such that:

Then, I substitute eq.(1) into eq. (2), and then take the derivative dX dn
. However, although I found
the derivative dX dn
such that,

the relation between X and n still UNCLEAR (what dominates the positive or the negative sign).
So, what I need is that the professor asked us that we should be able to use the lecture note (provided next pages) as a method to prove a definite sign for equation (4), which seems to be negative too as in the lecture note. In addition, after you get the full algebraic proof, please define in this problem the ‘Definition’, ‘Proposition’ and ‘Proof’.

  2 11 1 1 1
( ) ( ) ( ) ( ) 1 1
( )
1 1 1 ( 1) ( 1)
1
α σ σ σ σ α α
σ α
σ n δ n n n σ ndX δ δ dn α αAβ αAβ αA αA αAβ Aα Aα α β
n δ αAβ Aα Aα
    
   


                           
        
   
 
(4)

1 11 (1)
(1 ) (2)
σ α
α
n δY αAβ αA αA
X AY δ n Y


        
    
 
2 1 1
( ) because 0,1 ( )
1 0 (3) ( 1) α σ σ α α σ n ndY δ dn α αAβ αAβ αA αA                 
2

Lecture Note
What follows is not a full algebraic proof; it is rather a sketch of how to overcome the ambiguity which, at first sight, seems to exist in the sign of equation (4). You should be able to use this as a method to prove a definite sign for equation (4).
To understand the idea, consider the equation for X :

Intuitively, n  has three effects on X : there is a ‘direct effect’ at a given value of Y , and this is negative (we have already shown in (3) that / 0 dY dn ); while there are two ‘indirect effects’ which operate through the fall in Y .
Regarding the two ‘indirect effects’, we can see from the equation for X that they work in opposite direction: Y falls, and through α AY this lower X (holding all else constant); while through (1 ) δ n Y   , the fall in Y raises X (since 1 0 δ n    ). In summary, the two ‘indirect effects’ conflict with each other, so the sign of the ‘net indirect effect’ is unclear.
Let’s consider the sign of the ‘net indirect effect’ further.
We can write:

We can sketch X as a function of Y ( treating n as given), as follows:

where   Y is the value of Y which maximizes X .
(1 ) αX AY δ n Y      
( 1) α X AY n δ Y      
3

This shows that the ambiguity in the sign of the ‘net indirect effect’ stems from the fact that X is a function of Y is ‘hill-shaped’: For low values of Y , / 0 dX dY  ; while for high values of Y , / 0 dX dY  .
 we could overcome the ambiguity in the sign of the ‘net indirect effect’ if we know whether, in the initial, Y lies below the value   Y , or whether Y lies above   Y .
Now   Y is the value of Y such that:

So   Y  satisfies:

Next, we know that the actual value of Y is given by (1);

Compare equations (I) and (II). They are quite similar. If 1 n  , then for sure 1 1 σn δ n δ β      , since 1 β  . And if 1 n  , the same is true, since n is then > 1. Hence, we can say that, if 1 n ,

given that, on both sides of this, we have the same, decreasing, function of Y or   Y , it follows that:

In other words, the actual value of Y occurs on the positively sloped post of the curve overleaf.

Returning to our original argument, this proves that (provided 1 n  ), the ‘net indirect effect’ of n on X is unambiguously negative (recalling that n  causes Y  ). Given that the ‘direct effect’ of n  on X is also negative (see above), it then follows that the overall effect is also negative.
1