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Could someone prove why X and $Z+\Theta$ follow the same distribution with equations?

The text is taken from Ross A first course in probability

torgny
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  • Sure. But what exactly is it that you're stuck on? – Nap D. Lover Oct 04 '17 at 13:27
  • I dont get their reasoning. It would be better if someone could explain by using math signs if possible? Or try to explain it less subtle. – torgny Oct 04 '17 at 13:34
  • Ok but to be clear the proof asserts $Z+\Theta$ $\textbf{given}$ that $\Theta=\theta$ is distributed like $X$ given $\Theta=\theta$. You know that if $Z$ is standard normal then $a+bZ$ is Normal with mean $a$ and std dev $b$, right? Do you see how this answers your question? – Nap D. Lover Oct 04 '17 at 13:50
  • Is $\Theta$ a constant?It says that it is a normal random variable. It also says that Z is a standard normal variable. What do you mean by a+bZ? Is a a constant. I cant follow? – torgny Oct 04 '17 at 19:56
  • Do you not understand the phrase "given $\Theta=\theta$"? – Nap D. Lover Oct 04 '17 at 19:58
  • But they explicitly say that $\Theta=\theta$ is a variable with var=1 and Z also has var=1. How does then $Z+\Theta$ have var=1? – torgny Oct 04 '17 at 20:25
  • Please see my answer. To your most recent comment: since we treat $\Theta=\theta$ as a constant in the expression $Z+\Theta=Z+\theta$, AND since we know the variance of any constant is zero, we have that $var(Z+\theta)=var(Z)=1$. Does this help? – Nap D. Lover Oct 04 '17 at 20:31

1 Answers1

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So first of all, you are asking to prove that $Z+\Theta | \Theta=\theta$ is distributed like $X | \Theta=\theta$ and you are given by assumption the latter is distributed like a normal random variable with mean $\theta$ and variance $1$ and that $Z$ is a standard normal RV. Okay? So this means you need to show $Z+\Theta|\Theta=\theta$ is a normal random variable with mean $\theta$ and variance $1$.

Now, the first thing you do, is use the fact that you are given $\Theta=\theta$ and instead examine $Z+\theta$, treating $\theta$ as a constant. If is this confusing, you need to review conditional probability/conditional densities.

The second thing to do is either recall or learn the general fact that if $a,b$ are constants ($b\neq 0$), then $a+bZ$ is normally distirbuted with mean $a$ and standard deviation $b$ whenever $Z$ is a standard normal random variable.

Thus, by the above fact $Z+\theta$ is normally distirbuted with mean $\theta$ and variance $1$, proving your desired claim.

Now, how do we prove the general fact I am citing? Well two parts are exceptionally easy: $$E(a+bZ)=a+bE(Z)=a+b\cdot 0=a,$$ since $Z$ has mean $0$ and $E(\cdot)$ is a linear operator. Similarly, $$Var(a+bZ)=0+b^2var(Z)=b^2,$$ since $Z$ has variance $=1$. Thus, $Y=a+bZ$ has variance equal to $b^2$ and standard deviation equal to $b$. If these steps are confusing, then you need to review properties of expectation and variance.

All that is left is to show $Y=a+bZ$ has a normal distribution, then we're done since we just computed the mean and variance and they were as claimed. So, a typical method is to look at $$F_Y(y)=P(Y\leq y)=P(a+bZ\leq y)=P(Z\leq (y-a)/b)=F_Z(\tfrac{y-a}b),$$ then $f_Y(y)=\frac{1}{b} f_Z(\tfrac{y-a}{b}).$ Can you conclude from here?

Please comment if you have any questions or want further clarification.

Nap D. Lover
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