What is Distribution Function?

"Distribution Function" and "Cumulative Distribution Function" mean the same thing.

You would find $\displaystyle \int_{-\infty}^{x} u^3 \cdot f(u) \mathrm{d}u$

To remove the ambiguity; are you talking about a discrete distribution or a continuous one?

Actually nothing was mentioned specifically by CIE. I will check out some other A level exam board spec for a hint.

Yeah, I just had a look and it's ambiguous; I don't think looking at another board will tell you with confidence what they mean.

So, the two cases:

If you're talkijng about a discrete variable, then it's pretty obvious. If you know the distribution p(X = x) for X (e.g. for a die. p(X = k) = 1/6, for k = 1,..,.6), then if Y = X^2, then p(Y = j) will be the sum of all p(X=k) for k satisfying k^2 = Y. In other words, you find the values (*) of X that give the value of Y that you want and add them up.

(*) there may be 0, 1 or more possible values for X.

What I think is far more likely is that they're talking about continuous functions. In which case if X has a cumulative distribution function F (so that F(x) = P(X <= x)), then if Y = X^3, then obviously P(Y <= y) = P(X <= y^(1/3)) = F(y^(1/3)). So finding the cumulative function for Y is relatvely easy.

But what I think they may actually be getting at is what happens if you want to do the same thing with probability density functions. E.g. suppose X is uniformly distributed on [0,1] (so X has pdf p(X=x) = 1 (0 <= x < = 1), 0 otherwise. If Y = X^3, what is the pdf for Y?

To do this, you basically have to convert to cumulative functions and back. The cdf F(x) for X is P(X<=x) = x (assuming 0<=x<=1).. So the cdf G(y) for Y is P(Y<=y) = P(X <= y^(1/3)) = y^(1/3).

But the pdf for Y is just the derivative of the cumulative density function, so the pdf for Y is $\dfrac{d}{dy}y^{1/3} = 1/3 y^{-2/3}$.

The reason I think it's this is basically because there's actually "something to teach" with this. The first two options hardly seem worth an entry on the specification. I have definitely seen STEP questions where you need to do the pdf->cdf->pdf steps to get an answer (and I only really look at STEP questions, so that's not to imply it doesn't come up at A-level).

You sure about that integral? Doesn't look right to me.

Wiki also says "distribution function" is ambiguous:

https://en.m.wikipedia.org/wiki/Probability_distribution_function

To remove the ambiguity; are you talking about a discrete distribution or a continuous one?

That's "probability distribution function". "Distribution function" commonly refers to the CDF. See the first sentence on this page: https://en.wikipedia.org/wiki/Cumulative_distribution_function

I don't see what's wrong with the integral. $-\infty$ refers to the lower limit.

I stand corrected on distribution function, although I certainly wasn't aware that was the standard definition, and I'm unconvinced it's what is intended here.

What's your integral supposed to represent? And is f supposed to be the cumulative distribution function or the pdf?

Edit: I'm finding it really hard to think of any meaning for your integral that is going to work if we're considering the random variable uniformly distributed on [-2,-1], say (since u^3 will be -ve).