Mgf of a random variable with added constant

In summary, the author is trying to calculate the moment generating function of a r.v. Y=Z+c, but is having difficulty. They try to use an exponential function but it turns out to be incorrect. They then find a different way to calculate the function and are successful.
  • #1
Hey,

I have a pdf of a random variable Z given. I am being asked to calculate what the moment generating function of a r.v Y= Z + c will be where c is a constant in ℝ

I tried to calculate it in the following way:

[tex] \int^∞_0 e^{(z+c)t} f(z+c)dz[/tex] where [tex] f(z) [/tex] is an exponential pdf with parameter λ.

but it proved to be an unsuccessful method. Could anyone please show me the right direction? I know I could use Jacobian transformation but I'm sure there is an easier method.

Thank you in advance!
 
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  • #2
I wouldn't even mess around with the integral. Here is something I would try:

[itex] Y = Z + c [/itex] where [itex] Z ~ exp(\lambda) [/itex] and c is a constant. Then,

[itex]E[e^{tY}] = E[e^{t(Z+c)}] [/itex]

Now do you see what you might be able to do?
 
  • #3
Robert1986 said:
I wouldn't even mess around with the integral. Here is something I would try:

[itex] Y = Z + c [/itex] where [itex] Z ~ exp(\lambda) [/itex] and c is a constant. Then,

[itex]E[e^{tY}] = E[e^{t(Z+c)}] [/itex]

Now do you see what you might be able to do?

I think it definitely solves this problem! Now I can proceed with the rest of the exercise. Thank you Robert!
 
  • #4
You're most certainly welcome.

As a side note, this sort of thing is a rather valuable technique in prob/stat. That is, if you want to know about a certain RV, or a certain expectation, lots of times it is best to work it into some form you already know.
 
  • #5
WantToBeSmart said:
I think it definitely solves this problem! Now I can proceed with the rest of the exercise. Thank you Robert!

Of course, you would have gotten the same result had you used the correct f(z) dz in your integration, instead of your _incorrect_ f(z+c) dz.

RGV
 
  • #6
Ray Vickson said:
Of course, you would have gotten the same result had you used the correct f(z) dz in your integration, instead of your _incorrect_ f(z+c) dz.

RGV

Checked that and it was another mistake I was making. Thank you for pointing this out!
 

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