Random variate X follows a normal distribution

TomJerry
Messages
49
Reaction score
0
Question :
Random variate X follows a normal distribution with mean 0 and variance 1 i.e.X~N(0,1). Given Y = 2X + 4, find
i) E[Y]
ii) Var(Y)
iii) E[X^3]


Solution:

here E[X'] = 0 and V(X) = 1

i) E[Y] = E[2X+4] = 4 [Is this correct]

ii) Var(Y) = E[Y2] - [E(Y)]2

=E[Y2] -16

Now

E[Y2] = E[(2X+4)2]

= E[(4x2 + 16x + 16]

= 4 E[x2] + 16 E[x] + 16

= 4 E[x2] + 0 + 16

Stuck here ...how to evaluate E[x2]
 
Physics news on Phys.org
TomJerry said:
Question :
Random variate X follows a normal distribution with mean 0 and variance 1 i.e.X~N(0,1). Given Y = 2X + 4, find
i) E[Y]
ii) Var(Y)
iii) E[X^3]


Solution:

here E[X'] = 0 and V(X) = 1

i) E[Y] = E[2X+4] = 4 [Is this correct]

ii) Var(Y) = E[Y2] - [E(Y)]2

=E[Y2] -16

Now

E[Y2] = E[(2X+4)2]

= E[(4x2 + 16x + 16]

= 4 E[x2] + 16 E[x] + 16

= 4 E[x2] + 0 + 16

Stuck here ...how to evaluate E[x2]

I was able to figure out E[X^2] = 1 now i am stuck with getting the iii) question

E[X^3] how should i do this ..
 
You should post homework questions in the homework section.
If you know the mean is zero, the distribution is symmetric about zero, so ...
 
statdad said:
You should post homework questions in the homework section.
If you know the mean is zero, the distribution is symmetric about zero, so ...

Sorry for that , but since i have already posted it , could you please guide me on how to solve the last one

ie E[X^3] = ?
 
TomJerry said:
Sorry for that , but since i have already posted it , could you please guide me on how to solve the last one

ie E[X^3] = ?

Do you know the general definition of E[f(X)]? You are given the distribution so you should know the probability density function. Based on how E[f(X)] is defined, can you figure out how to calculate E[X^3]?
 
chiro said:
Do you know the general definition of E[f(X)]? You are given the distribution so you should know the probability density function. Based on how E[f(X)] is defined, can you figure out how to calculate E[X^3]?

Hey chiro,

I know that its mean is 0 that means xf(x) = 0 ...please help me with this...
 
TomJerry said:
Hey chiro,

I know that its mean is 0 that means xf(x) = 0 ...please help me with this...

As you probably know you have either discrete, continuous, or mixed (ie sometimes discrete and sometimes continuous) distributions.

The normal is continuous so let's focus on this, but the idea is easily seen with discrete when you grasp the core idea.

As you have said the mean is the integral over the whole domain ie (-infinity,infinity) which is

Integral [x . f(x) dx] over the whole real line (or the domain of the random variable, whatever is smaller).

The expectation of a function of x, where the function is g(x) is given by

Integral [g(x) . f(x) dx] over the whole real line

With your mean E[X] the g(X) is simply X which is why the mean is given by x . f(x).

So using the general formula, with a specific case of g(X) = X^3 we use formula

Integral [g(X) . f(X) dx] = Integral [x^3 . f(x)] dx over the whole real line.

f(x) is your pdf which you can get from a stats book, wiki site etc.

The rest is simply integration that you have learned in Calculus I and II.

You can actually find E[X^n] where n is a natural number using moments as well, but for now you need to stick to the basics and find things from first principles.
 
chiro said:
As you probably know you have either discrete, continuous, or mixed (ie sometimes discrete and sometimes continuous) distributions.


f(x) is your pdf which you can get from a stats book, wiki site etc.

The rest is simply integration that you have learned in Calculus I and II.

You can actually find E[X^n] where n is a natural number using moments as well, but for now you need to stick to the basics and find things from first principles.

Thanks ,
but my real concern is how will i get f(x) since i only know mean which is 0
so what i can do is f(x) = 0 and nothing else but not sure that the right way to go.
 
TomJerry said:
Thanks ,
but my real concern is how will i get f(x) since i only know mean which is 0
so what i can do is f(x) = 0 and nothing else but not sure that the right way to go.

All your "standard" distributions whether they are continuous (ie normal, gamma, student-t, etc) or discrete (ie binomial, geometric etc) have explicit formulas for the probability density function.

When you learn these probability functions, try to understand what assumptions have been made to derive the pdfs. Some distributions have different uses like the normal, which is used in the central limit theorem, or the limit of a sampling distribution, as well as something which conveniently explains a lot of natural phenomena.

Something like the binomial or geometric has assumptions which are used to derive the pdf. For example the binomial has N trials with only two outcomes. The big assumption being independence that you get from your probability axioms (ie P(A and B) = P(A)P(B)) helps define the pdf.

Like I said for standard distributions, any statistics book, math site (like Wolfram), or wiki page will have the pdf, and like I said each distribution has a specific purpose depending on the statistics and which area of statistics is involved. To get an idea of why the distribution is used again consult a stats book, ask your lecturer a question, look up the wiki page, or ask a question in the forums.

So to answer your question for f(x) look at:

http://en.wikipedia.org/wiki/Normal_distribution
 
  • #10
tomjerry, you need to show what YOU have tried on a particular homework problem: do not attempt to wheedle solutions from others.

If you understand the normal distribution, think again about the implication of \mu = 0.
 
Back
Top