1. ## Normal distribution?

$\displaystyle X\sim N(\mu,\sigma).$
$\displaystyle Y\equiv aX+b$.
What is $\displaystyle Y$'s distribution?
I guess $\displaystyle Y$ is still normally distributed, but what are $\displaystyle \mu_Y$ and $\displaystyle \sigma_Y$?

E.g. for $\displaystyle \sigma$: it's "intuitive" to think $\displaystyle \sigma_Y = |a|\sigma_X$, but how do you get the answer formally, with procedure? Is it via probability density function $\displaystyle p_X(x)$?

2. Originally Posted by courteous
I guess $\displaystyle Y$ is still normally distributed, but what are $\displaystyle \mu_Y$ and $\displaystyle \sigma_Y$?

E.g. for $\displaystyle \sigma$: it's "intuitive" to think $\displaystyle \sigma_Y = |a|\sigma_X$, but how do you get the answer formally, with procedure? Is it via probability density function $\displaystyle p_X(x)$?
Usually cumulative distributive function is used to compute distributions of transformation of random variables. You can also use the Jacobian approach.

I will start you off using the c.d.f $\displaystyle F_{Y}(y)$, where y is a real number:

$\displaystyle F_{Y}(y) = \text{Pr}(Y \leq y) = \text{Pr}(aX+b \leq y)$

What will you do next?

3. I can only manage 2 steps: $\displaystyle =Pr(X\leq \frac{y-b}{a})=\frac{1}{2}+\Phi(\frac{y-b}{a})$

4. Originally Posted by courteous
I can only manage 2 steps: $\displaystyle =Pr(X\leq \frac{y-b}{a})=\frac{1}{2}+\Phi(\frac{y-b}{a})$
Good! So how is a c.d.f related to a p.d.f?

5. Well, CDF is the sum of all the PDFs: $\displaystyle cdf(x)=\int_{-\infty}^x pdf(x)dx$.
I'm also looking at Wikipedia on change of variables, but I don't know how to make use of it.

I'm also guessing I shouldn't be doing $\displaystyle \int_{-\infty}^x \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$?

6. Originally Posted by courteous
Well, CDF is the sum of all the PDFs: $\displaystyle cdf(x)=\int_{-\infty}^x pdf(x)dx$.
I'm also looking at Wikipedia on change of variables, but I don't know how to make use of it.

I'm also guessing I shouldn't be doing $\displaystyle \int_{-\infty}^x \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$?
You should know that the derivative of the cdf gives the pdf.

7. $\displaystyle p_Y(y)=F_Y(y)'=\big(\frac{1}{2}+\Phi(\frac{y-b}{a})\big)'=0+\big(\frac{1}{2\pi}\larger{\int_0^{ \frac{y-b}{a}}}e^{-\frac{t^2}{2}}dt\big)'=\frac{1}{2\pi}...$

Now what?
How to relate $\displaystyle \frac{y-b}{a}$ with $\displaystyle t$?

8. Originally Posted by courteous
$\displaystyle p_Y(y)=F_Y(y)'=\big(\frac{1}{2}+\Phi(\frac{y-b}{a})\big)'=0+\big(\frac{1}{2\pi}\larger{\int_0^{ \frac{y-b}{a}}}e^{-\frac{t^2}{2}}dt\big)'=\frac{1}{2\pi}...$

Now what?
How to relate $\displaystyle \frac{y-b}{a}$ with $\displaystyle t$?
Actually now this question belongs to the Calculus forum :P

Repeating mr fantastic's words: You should know that $\displaystyle (F_X(x))' = f(x)$ where $\displaystyle F_X(x)$ is the c.d.f and $\displaystyle f(x)$ is the p.d.f of a random variable X.

So to solve the problem, I am expecting you to know the p.d.f of a normal distribution and chain rule of differentiation. Do you know these?

9. Originally Posted by courteous
$\displaystyle p_Y(y)=F_Y(y)'=\big(\frac{1}{2}+\Phi(\frac{y-b}{a})\big)'=0+\big(\frac{1}{2\pi}\larger{\int_0^{ \frac{y-b}{a}}}e^{-\frac{t^2}{2}}dt\big)'=\frac{1}{2\pi}...$

Now what?
How to relate $\displaystyle \frac{y-b}{a}$ with $\displaystyle t$?
The derivative is with respect to y. Use the chain rule and the Fundamental Theorem of Calculus to get it.

10. Isn't what's inside the integral (that I wrote above) "the p.d.f of a normal distribution"?
Notation: I've mistakenly used $\displaystyle p_Y(y)$ for $\displaystyle f(y)$.
Chain rule is $\displaystyle \frac{da}{db}=\frac{da}{dc}\frac{dc}{db}$.

I haven't come upon such a problem before, so I'm not "used to it", as von Neumann would say.

Was I even on the right track in my previous post?

11. Originally Posted by courteous
Isn't what's inside the integral (that I wrote above) "the p.d.f of a normal distribution"?
Notation: I've mistakenly used $\displaystyle p_Y(y)$ for $\displaystyle f(y)$.
Chain rule is $\displaystyle \frac{da}{db}=\frac{da}{dc}\frac{dc}{db}$.

I haven't come upon such a problem before, so I'm not "used to it", as von Neumann would say.

Was I even on the right track in my previous post?
You are on the right track. However you are not using the fact that derivative of a c.d.f is the p.d.f. Your integral along with the $\displaystyle \frac12$ term is the c.d.f of a normal distribution, i.e.

$\displaystyle F_Y(y) = F_X\left(\frac{y-b}{a}\right) \implies p_Y(y) = (F_Y(y))' =\left(F_X\left(\frac{y-b}{a}\right)\right)' = \text{???}$

Alternatively,

$\displaystyle \big(\frac{1}{2\pi}\larger{\int_0^{\frac{y-b}{a}}}e^{-\frac{t^2}{2}}dt\big)' = \dfrac{d}{dy}\big(\frac{1}{2\pi}\larger{\int_0^{\f rac{y-b}{a}}}e^{-\frac{t^2}{2}}dt\big)$

I was telling you how to proceed. You can use the fundamental theorem of calculus to differentiate integrals with respect to a variable that appears in the limit of the integral. In the above expression 'y' appears as a limit of integration. So you have to use the fundamental theorem of calculus.

12. Originally Posted by Isomorphism
$\displaystyle F_Y(y) = F_X\left(\frac{y-b}{a}\right) \implies p_Y(y) = (F_Y(y))' =\left(F_X\left(\frac{y-b}{a}\right)\right)' = \text{???}$
$\displaystyle \displaystyle{ =\frac{d}{dy}\big(\int_0^{\frac{y-b}{a}}\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx\big)=\frac{1}{\sigma\sqrt{2\ pi}}\frac{d}{dy}\big(\int_0^{\frac{y-b}{a}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx\big) \overbrace{=}^{A} \frac{1}{\sigma\sqrt{2\pi}} \frac{d}{dy}\big(h(g(y))\big)}= }$

$\displaystyle \displaystyle{ = \frac{1}{\sigma\sqrt{2\pi}} h'\big(g(y)\big)g'(y) \overbrace{=}^{B} \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(\frac{y-b}{a}-\mu)^2}{2\sigma^2}} \frac{1}{a} = \frac{1}{(a\sigma)\sqrt{2\pi}} e^{-\frac{(y-(a\mu+b))^2}{2(a\sigma)^2}}}$

$\displaystyle \implies Y\sim N(a\mu +b, |a|\sigma)$ Is everything correct?

A: Let $\displaystyle h(y)=\int_0^y e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx$ and $\displaystyle g(y)=\frac{y-b}{a}$.

B: $\displaystyle h'(y)=\frac{d}{dy}\big(h(y)\big)=\frac{d}{dy}\big( \int_0^y e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx\big)\overbrace{=}^{FTC} e^{-\frac{(y-\mu)^2}{2\sigma^2}}$. FTC = Fundamental theorem of calculus.

13. Originally Posted by courteous
$\displaystyle \displaystyle{ =\frac{d}{dy}\big(\int_0^{\frac{y-b}{a}}\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx\big)=\frac{1}{\sigma\sqrt{2\ pi}}\frac{d}{dy}\big(\int_0^{\frac{y-b}{a}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx\big) \overbrace{=}^{A} \frac{1}{\sigma\sqrt{2\pi}} \frac{d}{dy}\big(h(g(y))\big)}= }$

$\displaystyle \displaystyle{ = \frac{1}{\sigma\sqrt{2\pi}} h'\big(g(y)\big)g'(y) \overbrace{=}^{B} \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(\frac{y-b}{a}-\mu)^2}{2\sigma^2}} \frac{1}{a} = \frac{1}{(a\sigma)\sqrt{2\pi}} e^{-\frac{(y-(a\mu+b))^2}{2(a\sigma)^2}}}$

$\displaystyle \implies Y\sim N(a\mu +b, |a|\sigma)$ Is everything correct?

A: Let $\displaystyle h(y)=\int_0^y e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx$ and $\displaystyle g(y)=\frac{y-b}{a}$.

B: $\displaystyle h'(y)=\frac{d}{dy}\big(h(y)\big)=\frac{d}{dy}\big( \int_0^y e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx\big)\overbrace{=}^{FTC} e^{-\frac{(y-\mu)^2}{2\sigma^2}}$. FTC = Fundamental theorem of calculus.

Perrrrfect!!

14. Thank you both, Isomorphism and Mr. Fantastic, for your guidance (and patience) ... I am especially grateful that you didn't show the whole solution at once.

15. ## The Jacobian approach

$\displaystyle X\sim N(\mu,\sigma).$
$\displaystyle Y\equiv aX+b$.
What is $\displaystyle Y$'s distribution?
Originally Posted by Isomorphism
Usually cumulative distributive function is used to compute distributions of transformation of random variables. You can also use the Jacobian approach.
Isomorphism, would you show me "the Jacobian approach", please? (Other members are welcomed as well!)

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