Gujarati Chapter 3

[srinu2singaraju]

Hi david,

In Gujarati
Chapter 2
F(Y/X) = F(XY)/F(X) = Joint probability of X and Y / Marginal probability of X

Example : 2.17
To calculate the probability that 4 printers were sold, knowing 4 PCs were sold can be calculated as
From Table 2.4:- F(Y=4/X=4) = 0.15/0.32 which is 0.47

Chapter 3
In Example 3.9:

E(Y/X=2) has calculated thru ∑Y F(Y/X=2)

My doubt is why should we multiply with Y values?

Y can take any value between 0 and 4 for a given value of X
so the probability of happening Y given X=2 should be

F(Y=0/X=2)+F(Y=1/X=2)+ F(Y=2/X=2)+F(Y=3/X=2)+ F(Y=4/X=2)

This is what my understanding,

But in the book it was mentioned as

F(Y=1/X=2)+ 2 F(Y=2/X=2)+3 F(Y=3/X=2)+ 4 F(Y=4/X=2)

Both the cases looks are similar to me first case we need to find the probability of exactly 4 printers sold, second case we need find all the possible values of Y.
Please clarify why we have to multiply with 2,3,4…

srinivas

 

[David Harper CFA FRM]

Hi srinivas,

(FYI, I refer to XLS @ http://www.bionicturtle.com/premium/spreadsheet/2.a.1._bivariate_pdf/)

If we take example 3.9 but FIRST answer the conditional probablity (PMF) question that is asked in 2.17, we ask the conditional probability question: What is P(sell 2 printers | sold 2 PCs)?

Answer = P (sell 2 printers, sell 2 PCs) / P (sold 2 PCs)
= joint / unconditional = 0.1 = .24 = 0.41667

These are "merely" probabilities and given the five outcomes {0...4}, as a probability, they must sum to 1.0 in the vertical column:
P (sell 0 printer | sold 2 PCs) = (4/200)/(48/200) = 4/48
P (sell 1 printer | sold 2 PCs) = 12/48
P (sell 2 printer | sold 2 PCs) = 20/48
P (sell 3 printer | sold 2 PCs) = 10/48
P (sell 2 printer | sold 2 PCs) = 2/48

Now we move to the actual 3.9 question which is different because it asks for the conditional EXPECTED VALUE, to get that we need to multiply the conditional probabilities by the respective number of printers sold. This is a weighted average as your language reflects...it is similar to asking "what is the expected value of rolling a six-sided die?" Note the answer is given by:

P(roll 1) * outcome of 1 + P(roll 2) * outcome of 2 ... + P(roll 6) * outcome 6
= 1/6*1 + 1/6*2 + ... + 1/6*6
= 1/6*(1+2+...+6) = 21/6 = 3.5 is the expected value of rolling a single die
(it's a long way around but just trying to show...)

(... this is related to a common confusion re: continuous distribution like the normal: f(x) is not the probability, it is only the vertical height of the Y-axis. The probability is the area under the curve so it can be approximated with dx*f(x) ... this helps to explain why some of our functions produce f(x) > 1.0)

Hope that helps, David

 

[srinu2singaraju]

Thanks.. david.

You have explained well. But still i have some doubts on the use of these equations 2.24 and 3.34 (from Gujarati)
If you can give any example where i can apply the above equations for better clarity.
My issue is in any given question, where should i use Eq:2.24 and 3.34

srinivas