Statistics: variable transformation proof?

[Nikitin]

Homework Statement

Ok this might be a stupid question, but:

https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-frc3/t31/q77/s720x720/10001118_10202561443653973_1625797585_o.jpg
Why is this the case? I think for all of this to be right, then the assumption of $Y=u(X) \Leftrightarrow y=u(x)$. But how do I prove this?

The Attempt at a Solution

An attempt: OK, assume $X$ has a probability distribution given by $f(x)$, and $Y=2X$ has a probability distribution of $g(x)$. Then if $Y = u(X)$, with w being the inverse one-on-one function of u, $P(X=x) = P(w(Y)=x) = P(Y=u(x))=P(Y=y=u(x))$.

This is as far as I got, but now I am confused. What to do?

 

[micromass]

You're almost there. Note by definition that

g(y) = P\{Y = y\}

and you must prove that

g(y) = f(w(y))

which is by definition

f(w(y)) = P\{X = w(y)\}

So you must somehow find why

P\{Y = y\} = P\{X=w(y)\}

 

[Nikitin]

OK, I see. Thanks for the help :)

 

[Ray Vickson]

Homework Statement

Ok this might be a stupid question, but:

https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-frc3/t31/q77/s720x720/10001118_10202561443653973_1625797585_o.jpg
Why is this the case? I think for all of this to be right, then the assumption of $Y=u(X) \Leftrightarrow y=u(x)$. But how do I prove this?

The Attempt at a Solution

An attempt: OK, assume $X$ has a probability distribution given by $f(x)$, and $Y=2X$ has a probability distribution of $g(x)$. Then if $Y = u(X)$, with w being the inverse one-on-one function of u, $P(X=x) = P(w(Y)=x) = P(Y=u(x))=P(Y=y=u(x))$.

This is as far as I got, but now I am confused. What to do?

You already got the answer below, but just to clarify: the one-to-one assumption is crucial. If you had a non-monotone function, such as $u(x) = x^2$, and if the range of $X$ included both positive and negative values, then the mapping $Y = u(X)$ might not be 1:1, and so you might need a more complicated evaluation of $P(Y = y)$.