STEP Prep Thread 2020

@Drogo Baggins Just to add another obvious typo,

Can anyone tell me what this actually means? (Screenshot from spec attached)

I remember calculating the mean and variance of X² knowing that of X for discrete and continuous distributions but I don't ever recall using a cdf to work out a pdf of a related variable. How is this even done?

Can anyone tell me what this actually means? (Screenshot from spec attached)

I remember calculating the mean and variance of X² knowing that of X for discrete and continuous distributions but I don't ever recall using a cdf to work out a pdf of a related variable. How is this even done?

Suppose $X$ is a random variable with probability density function $f(x)$ and cumulative distribution function $F(x)$, such that $\dfrac{d}{dx}F(x) = f(x)$.

Introduce a random variable $Y = \gamma(X)$ where $\gamma$ is some nice function with the required properties to justify what I am about to write (i.e. inverse exists, monoticity, etc...), and it represents a related random variable $Y$ to $X$ through a function $\gamma$. The cumulative distribution function of this is

$G(y) = P[Y \leq y] = P[\gamma(X) \leq y] = P[X \leq \gamma^{-1} (y)] = F[\gamma^{-1}(y)]$

so we have related G to F in some way, and so to obtain the density function you just differentiate with respect to y;

$g(y) = \dfrac{d [\gamma^{-1} (y)]}{dy} f[\gamma^{-1}(y)]$

Correct, but be aware that some of the simple examples that might be asked do not meet your definition of a "nice" function!

This one's pretty funny 🤣

Why is it funny?

Of all the possible maths typos ...

1=0

xD

Suppose $X$ is a random variable with probability density function $f(x)$ and cumulative distribution function $F(x)$, such that $\dfrac{d}{dx}F(x) = f(x)$.

Introduce a random variable $Y = \gamma(X)$ where $\gamma$ is some nice function with the required properties to justify what I am about to write (i.e. inverse exists, monoticity, etc...), and it represents a related random variable $Y$ to $X$ through a function $\gamma$. The cumulative distribution function of this is

$G(y) = P[Y \leq y] = P[\gamma(X) \leq y] = P[X \leq \gamma^{-1} (y)] = F[\gamma^{-1}(y)]$

so we have related G to F in some way, and so to obtain the density function you just differentiate with respect to y;

$g(y) = \dfrac{d [\gamma^{-1} (y)]}{dy} f[\gamma^{-1}(y)]$

I'm trying this with a given pdf for X and trying to work out the pdf for X². In the end I'm left with a function in terms of √y. At this point, do I just substitute in x = √y ?

Picture attached

Maybe I chose an unfortunate example because the pdf of X² in the case I chose ends up being sinx/4x which can't be integrated back in terms of x without using the Si(x) function which I'm not familiar with or should be aware of for STEP. So I am unable to determine if the area of this function is still 1 when evaluated between 0 and pi or if it has new bounds it needs to be evaluated from such that the area under the described pdf is still 1.

(Using a computer approximation, it outputs 0.463 as the area between 0 and pi - have I done something wrong in carrying out the process?)

@Drogo Baggins Just to add another obvious typo,

You should integrate your pdf result with sqrt(y) in it between 0 and pi^2. See if you can spot your error.

Can anyone tell me what this actually means? (Screenshot from spec attached)

I remember calculating the mean and variance of X² knowing that of X for discrete and continuous distributions but I don't ever recall using a cdf to work out a pdf of a related variable. How is this even done?

Seems like I get 1 with those bounds:

Maybe this means that the bounds of X² is that of X squared (would make sense I guess).

I feel like I've made some sort of mistake neglecting the fact that there could be a negative number for the square root as well such that if P(X²≤y) then this is the same as P(-√y≤X≤√y) but I don't know to what extent this has affected my answer (if it has made any difference at all)

Thanks! What I don't understand is that 1 and 0 are not that close on a keyboard?

Look at the sub you did... If you want to integrate with respect to x (which is what you were doing before) then you need to relate x with y somehow ... But we already have the relation!

x^2 = y

So your error in integration going into Si(x) was to do with the fact that you were neglecting a factor of dy/dx in the integrand which comes from using the substitution.

Anyway, you are allowed to take the positive root when square rooting because you know that x is positive.

If it was negative, you would need the negative sign on the root.

If x is a mixture, then it gets a bit more complicated and you need to consider both cases.

Thanks! What I don't understand is that 1 and 0 are not that close on a keyboard?