Constant times a random variable question

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A random variable X follows a certain distribution. Now say I multiply the random variable X by a constant a. Does the new random variable aX follow the same distribution as X?
 

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  • #2
EnumaElish
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Well, let's take it through the basics. If X ~ FX, then FX(x) = Prob{X < x} for x < X < [itex]\bar x[/itex]. Let Y = aX for ax < Y < a[itex]\bar x[/itex]; then FY(y) = Prob{Y < y} = Prob{aX < y} = Prob{X < y/a} = FX(y/a). Therefore FY(y) = FX(y/a) for ax < Y < a[itex]\bar x[/itex].

Is FY the identical distribution as FX? No. Does it belong to the same family as FX? Yes, up to a scaling factor.
 
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  • #3
EnumaElish
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Note: in the previous post I assumed a > 0. If a < 0, then FY may not even belong to the same family as FX.

The case of a = 0 is trivial, but you should bear that in mind, too.
 
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we know that if X is normally distributed, then so cX for any nonzero real number c.
also X + d is normally distributed, for any real number d.
can anyone please show me the proof?thanks
 
  • #5
chiro
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A random variable X follows a certain distribution. Now say I multiply the random variable X by a constant a. Does the new random variable aX follow the same distribution as X?
If you want a systematic way to figure this out, use moment generating functions and if the structure of the mgf is the same as the unscaled distributions mgf, then you know that the distribution doesn't change and you can see how the parameters of the distribution have changed.
 
  • #6
chiro
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we know that if X is normally distributed, then so cX for any nonzero real number c.
also X + d is normally distributed, for any real number d.
can anyone please show me the proof?thanks
Like I said to the previous poster, use moment generating functions. Let U = cX + d. Use E[e^(tU)] be the mgf of U and you will find that this will give the mgf of a normal distribution with different parameters which will tell you what scaling and translating a normal does to its parameters. (Translating adds to the mean, scaling changes the variance by multiplying it by c^2).
 

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