Am I doing this right?

[versed dagger]

so far this is what I have

Y = -ln(X)

inverse would be:

X = e^(-Y) and the derivative of that would be dX/dY = -e^(-Y)

but now when it comes to the bounds we have

0 < X < 1 => 0 < e^(-Y) < 1
I will have to take ln of both sides
so it will be ln(0) < -Y < ln(1) right? but ln(0) is not defined
so what should I do
do I just treat it as neg inf
so, -inf < -Y < 0 => 0 < Y < inf will be the bound?

If so then the pdf of Y would be

f_X(x) |dX/dY| = f_X(e^-Y) * e^(-Y) = 4(e^-Y)^3 * e^-Y = 4e^(-4Y) so the ans would be IV?

[versed edge]

1. Ask your question and show the work you've done so far. If you've posted a screenshot of a question, specify which part you need help with.
2. Wait patiently for a helper to come along.
3. Once someone helps you, say thank you and close the thread with:

   +close
   

4. Feel free to nominate the person for helper of the week in #helper-nominations
5. Do not ping the mods, unless someone is breaking the rules.
6. If you're happy with the help you got here, and the server overall, you can contribute financially as well:

[versed dagger]

🕊️

[tender magnet]

You can try with a test function

[versed dagger]

what does that mean

[tender magnet]

For any $\phi \geq 0$ measurable:
\begin{align*}
E[\phi(Y)] &= E[\phi(-\ln(X))] \\
&= \int_0^{1} \phi(-\ln(x)) f_X(x) dx \\
&= \int_{0}^{1} \phi(-\ln(x)) \cdot 4x^3 dx
\end{align*}

[jolly canopy]

Rion

[versed dagger]

but is my reasoning incorrect

[tender magnet]

Now you can do a variable substitution: $y = -\ln(x)$, which yields $dy = \frac{-dx}{x}$

[jolly canopy]

Rion

[versed dagger]

for IV

[tender magnet]

Oh

I just realized the reasoning you sent was kind of a test function but kinda hidden, so I was confused

the overall reasoning is correct

[versed dagger]

lastly

could you explain this

since it says X^2 = 1 we rlly are only looking for X = -1 and X = 1 here right

[tender magnet]

Well what's a generating function

[versed dagger]

i dont rlly know

[tender magnet]

$M_X(t) = E[e^{tX}]$

[jolly canopy]

Rion

[versed dagger]

wait so lets say for P(X = -2) it would be 1/4?

[tender magnet]

Well, hold your horses

That is indeed true, but

It is not possible to directly guess the distribution of $X$ given its MGF

[jolly canopy]

Rion

[tender magnet]

However, you can say that a distribution is uniquely determined by the MGF

[versed dagger]

oh

[tender magnet]

So you can make a random variable with the same MGF, and say that by extension, X has the same distribution

So that will be why you can indirectly guess the probabilities

[versed dagger]

in that case for

P(X = 1) = 1/6 and P(X = -1) = 1/4 right

[tender magnet]

yes

[versed dagger]

damn so then its

1/6 / (1/6 + 1/4) thats it right

the answer

bcz P(X = 1 | X^2 = 1) = P(X > 0 | X^2 = 1) here

= 2/5

0.4

[tender magnet]

Should be the right reasoning at least

[versed dagger]

bett ty

lightwork

+close