Variance of Continous Random Variable

  • Thread starter Lancelot59
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  • #1
634
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I have the continous random variable Y, defined such that:

[tex]
Y=3X+2
[/tex]
and the PDF of x is zero everywhere but:
[tex]f(x)=\frac{1}{4}e^{\frac{-x}{4}}, x>0[/tex]

I correctly got the mean like so:
[tex]\mu=E(h(x))=\int^{\infty}_{0} h(x)f(x)[/tex]
and evaluated it to be 10.

I am unsure as to how I go about getting the variance now. In previous problems I use the usual formula of:

[tex]\sigma=\int^{b}_{a} (x-\mu)^{2}f(x)[/tex]

or the shortcut of:
[tex]\sigma^{2}=E[X^{2}]-\mu^{2}[/tex]

However I'm unsure what to do in the case where the random variable is dependent on another with a given PDF function for continuous random variables.
 

Answers and Replies

  • #2
62
6
Start with substiting in the expression for Y into [itex]Var(Y)[/itex]

Seperate out [itex]Var(X)[/itex]
 
  • #3
634
1
Start with substiting in the expression for Y into [itex]Var(Y)[/itex]

Seperate out [itex]Var(X)[/itex]

I'm not quite sure how to put that together. To start, I have:
[tex]Var(Y)=\int^{\infty}_{0} (y-10)^2*f(y)[/tex]
 
  • #4
62
6
I would start with [itex]Var(Y)=Var(3X+2)=Var(3X)+Var(2)[/itex]
 
  • #5
62
6
I'm not quite sure how to put that together. To start, I have:
[tex]Var(Y)=\int^{\infty}_{0} (y-10)^2*f(y)[/tex]

If you want to use this form then [itex]\mu_Y=E(Y)=E(3X+2)=3E(X)+2[/itex]
 
  • #6
634
1
If you want to use this form then [itex]\mu_Y=E(Y)=E(3X+2)=3E(X)+2[/itex]

I see. So all I need to do is then find the expected value of the function defining the PDF of x?
 
  • #7
Ray Vickson
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I have the continous random variable Y, defined such that:

[tex]
Y=3X+2
[/tex]
and the PDF of x is zero everywhere but:
[tex]f(x)=\frac{1}{4}e^{\frac{-x}{4}}, x>0[/tex]

I correctly got the mean like so:
[tex]\mu=E(h(x))=\int^{\infty}_{0} h(x)f(x)[/tex]
and evaluated it to be 10.

I am unsure as to how I go about getting the variance now. In previous problems I use the usual formula of:

[tex]\sigma=\int^{b}_{a} (x-\mu)^{2}f(x)[/tex]

or the shortcut of:
[tex]\sigma^{2}=E[X^{2}]-\mu^{2}[/tex]

However I'm unsure what to do in the case where the random variable is dependent on another with a given PDF function for continuous random variables.

Neither X nor Y has mean 10, so you did something wrong. What is the mean of X? How is the mean of Y related to the mean of X? What is Var(X)? How is Var(Y) related to it?

RGV
 
  • #8
634
1
Neither X nor Y has mean 10, so you did something wrong. What is the mean of X? How is the mean of Y related to the mean of X? What is Var(X)? How is Var(Y) related to it?

RGV

10 is what I got, and it matches the answer in the textbook...
 
  • #9
jbunniii
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10 is what I got, and it matches the answer in the textbook...
I calculated E[X] = 4. Please show how you got 10.
 
  • #10
Ray Vickson
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10 is what I got, and it matches the answer in the textbook...

Well, EX (the mean of X) is not 10, and EY (the mean of Y) is not 10. However, E(3*X - 2) = 10, so maybe you just typed the question incorrectly, and maybe you really meant EY when you seemed to be talking about EX.

RGV
 
  • #11
634
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I calculated E[X] = 4. Please show how you got 10.

Here is the integral I solved in order to get the mean:

[tex]\mu=\int^{\infty}_{0} \frac{1}{4}e^{\frac{-x}{4}}(3x-2)dx=10[/tex]

from the identity where:

[tex]E[h(x)]=\int^{\infty}_{-\infty} h(x)f(x) dx[/tex]

Anyhow, it turns out that in solving the variance I made an arithmetical mistake, and I did manage to get the correct answer. Thanks for the help!
 
  • #12
Ray Vickson
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Here is the integral I solved in order to get the mean:

[tex]\mu=\int^{\infty}_{0} \frac{1}{4}e^{\frac{-x}{4}}(3x-2)dx=10[/tex]

from the identity where:

[tex]E[h(x)]=\int^{\infty}_{-\infty} h(x)f(x) dx[/tex]

Anyhow, it turns out that in solving the variance I made an arithmetical mistake, and I did manage to get the correct answer. Thanks for the help!

OK, so as I suspected, you computed EY = E(3X-2), but your question originally said Y = 3X+2!

BTW: if 'a' and 'b' are constants, Var(AX+b) = a^2 Var(X) [that is, the constant term 'b' drops out and the factor 'a' gets squared]. Since Var(X) = 4^2 = 16, we have Var(Y) = 3^2 * 16 = 144. This is a lot easier than computing
[tex] \int_0^{\infty} (y-10)^2 f_Y(y) \, dy.[/tex]

RGV
 

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