Beyond first-year and random math/stats questions/discussion

[RuiAce]

The question seems really weird and no one got the answer as of now lmao. It's a practice question from a prob and stats unit I'm doing now.

Still don't really have any suggestion. Although, the only hint that I can really suggest is that since it's an approximation, maybe think Taylor series?

Please let me know if you manage to obtain a full solution though

(Note: There's a proof on stack for E[Y] that doesn't use the MGF but uses Taylor series to jump straight into it)

[Springyboy]

The question seems really weird and no one got the answer as of now lmao. It's a practice question from a prob and stats unit I'm doing now.

I’m confused on this as well. Could you please help us out @Rui

My tutor used Taylor’s series of approximations but it was really vague, so if possible can you explain it out

EDIT:
This is the question btw

\begin{array} { l } { \text { Assume that } X \text { has a Poisson distribution with rate parameter } \lambda . \text { If } } \\ { Y = \sqrt { X } , \text { using moment generating functions or otherwise, show that } } \\ { E [ Y ] \approx \sqrt { \lambda } - 1 / ( 8 \sqrt { \lambda } ) \text { and } V a r [ Y ] \approx 1 / 4 \text { where the symbol } \approx \text { means } } \\ { \text { approximately equals', i.e. these formulas provide approximations to } } \\ { \text { the mean and variance of } Y } \end{array}

EDIT 2:

Got help and worked it out, have a look here:

https://math.stackexchange.com/questions/2908283/finding-ex-and-varx-of-poisson-distribution-where-y-sqrtx

[LifeisaConstantStruggle]

Still don't really have any suggestion. Although, the only hint that I can really suggest is that since it's an approximation, maybe think Taylor series?

Please let me know if you manage to obtain a full solution though

(Note: There's a proof on stack for E[Y] that doesn't use the MGF but uses Taylor series to jump straight into it)

Yeah we got our solutions today, it was exactly Taylor series approximations. The lecturer revised the question like 3 days ago so we're not restricted to only using MGFs. Thanks though!

Despite that the question still seemed really weird because Taylor series isn't taught in any preceding units leading up to this one (it isn't taught in Victorian high schools as well) unless if you count a maths unit that not everyone took before this unit.

[RuiAce]

Basically yeah this was the nicest approach I've been able to find thus far. The Delta method proof is quite interesting though - it makes perfect sense but I guess I just haven't used it enough in my life to really think about diving into it.

I guess at my university a more formal stats unit isn't taught until second year, so they've all seen Taylor series regardless before jumping into the course. But if they're gonna jump into theoretical statistics while risking no proper understanding of first year level mathematics then that's kinda up to them really. So long as it doesn't show up in the finals.

Pretty much, yeah, I think the MGF was one of the worst ideas for such a question. Taylor series are typically a go-to option for approximations involving smooth functions just because of how slick they are. (Although will wager that the Delta method is equally nice.)

[Springyboy]

Hey Rui,

Can you just help me out with integrating this function, I tried to use substitution but it keeps jamming up




It's for a Chebyshev's inequality question, but the RHS for Chebyshev's inequality is pretty straightforward to calculate so I just need help with doing this.
Thanks,

James

[RuiAce]

Hey Rui,

Can you just help me out with integrating this function, I tried to use substitution but it keeps jamming up




It's for a Chebyshev's inequality question, but the RHS for Chebyshev's inequality is pretty straightforward to calculate so I just need help with doing this.
Thanks,

James

I mean, given how low that power is I would've just bashed it with the binomial theorem.
\begin{align*}\int_\mathbb{R} f_X(x)\,\mathrm{d}x &= \int_0^1 630\left(x^8 - 4x^7 + 6x^6 -4x^5 + x^4\right)\mathrm{d}x\\ &= 630\left( \frac{x^9}{9} - \frac{x^8}{2} + \frac{6x^7}{7} - \frac{2x^3}{3} + \frac{x^5}{5} \right) \bigg|_0^1\\ &= 1 \end{align*}
Although that's also because I have class soon so I don't have enough time to think up a more clever way right now

EDIT: Or actually, you could note that \( \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)} x^{a-1} (1-x)^{b-1} \mathbb{I}\{0 < x < 1 \} \) is the density of a \( \mathsf{Beta}(a,b) \) distribution, in which you can then state that \( \int_0^1 x^4(1-x)^4\mathrm{d}x = \frac{\Gamma(5)^2}{\Gamma(10 )} \). This can be evaluated by using the key property that \( \Gamma(n) = (n-1)! \)

[Springyboy]

I mean, given how low that power is I would've just bashed it with the binomial theorem.
\begin{align*}\int_\mathbb{R} f_X(x)\,\mathrm{d}x &= \int_0^1 630\left(x^8 - 4x^7 + 6x^6 -4x^5 + x^4\right)\mathrm{d}x\\ &= 630\left( \frac{x^9}{9} - \frac{x^8}{2} + \frac{6x^7}{7} - \frac{2x^3}{3} + \frac{x^5}{5} \right) \bigg|_0^1\\ &= 1 \end{align*}
Although that's also because I have class soon so I don't have enough time to think up a more clever way right now

Legend cheers