How Can a Frog's Jumping Pattern Lead to Normal Distribution Behavior?

[FeDeX_LaTeX]

Homework Statement

Let X be a random variable with a Laplace distribution, so that its probability density function
is given by

f(x) = \frac{1}{2}e^{-|x|}

Sketch f(x). Show that its moment generating function M_X (θ) is given by

M_{X}(\theta) = \frac{1}{1 - \theta^2}

and hence find the variance of X.

A frog is jumping up and down, attempting to land on the same spot each time. In fact, in
each of n successive jumps he always lands on a fixed straight line but when he lands from the ith jump (i = 1 , 2 , . . . , n) his displacement from the point from which he jumped is X_i cm, where X_i has the Laplace distribution described above. His displacement from his starting point after n jumps is
Y cm, so that Y = \sum_{i=1}^{n} X_{i}.

Each jump is independent of the others.

Obtain the moment generating function for \frac{Y}{\sqrt{2n}} and, by considering its logarithm, show that this moment generating function tends to e^{\frac{1}{2}x^{2}} as n → ∞.

Given that e^{\frac{1}{2}x^{2}} is the moment generating function of the standard Normal random variable, estimate the least number of jumps such that there is a 5% chance that the frog lands 25 cm or more from his starting point.

Homework Equations

M_{X}(t) = E(e^{tX}) = \int_{-\infty}^{\infty} e^{tx}f(x)dx

The Attempt at a Solution

I've sketched f(x), which looks like the graph of \frac{1}{2}e^x for x < 0 and \frac{1}{2}e^{-x} for x > 0.

I've found the moment generating function, and deduced that it has mean 0 and variance 2.

However, I'm unable to obtain the moment generating function for \frac{Y}{\sqrt{2n}}. The mark scheme says this:

If T = \frac{Y}{\sqrt{2n}}, then M_{T}(\theta) = E(e^{T\theta}) = E(e^{\theta \sum \frac{X_{i}}{\sqrt{2n}}}) = \prod_{i=1}^{n}E(e^{\frac{\theta}{\sqrt{2n}}X_{i}}) = \left( 1 - \frac{\theta^{2}}{2n} \right)^n

I understand everything up until where the last part; how are they turning that product into that neat (1 - theta^2 / 2n)^n term? My approach was to say that all the X_i have the same distribution, so every term in the product is the same, and you get this:

\left( \frac{1}{\sqrt{2n}}M_{X}(\theta) \right)^n = \left(\frac{1}{\sqrt{2n}(1 - \theta^{2})} \right)^n

but this is clearly not equivalent to their answer. What have I done wrong here and what have they done to collapse their product into something so simple?

 

[Ray Vickson]

Homework Statement

Let X be a random variable with a Laplace distribution, so that its probability density function
is given by

f(x) = \frac{1}{2}e^{-|x|}

Sketch f(x). Show that its moment generating function M_X (θ) is given by

M_{X}(\theta) = \frac{1}{1 - \theta^2}

and hence find the variance of X.

A frog is jumping up and down, attempting to land on the same spot each time. In fact, in
each of n successive jumps he always lands on a fixed straight line but when he lands from the ith jump (i = 1 , 2 , . . . , n) his displacement from the point from which he jumped is X_i cm, where X_i has the Laplace distribution described above. His displacement from his starting point after n jumps is
Y cm, so that Y = \sum_{i=1}^{n} X_{i}.

Each jump is independent of the others.

Obtain the moment generating function for \frac{Y}{\sqrt{2n}} and, by considering its logarithm, show that this moment generating function tends to e^{\frac{1}{2}x^{2}} as n → ∞.

Given that e^{\frac{1}{2}x^{2}} is the moment generating function of the standard Normal random variable, estimate the least number of jumps such that there is a 5% chance that the frog lands 25 cm or more from his starting point.

Homework Equations

M_{X}(t) = E(e^{tX}) = \int_{-\infty}^{\infty} e^{tx}f(x)dx

The Attempt at a Solution

I've sketched f(x), which looks like the graph of \frac{1}{2}e^x for x < 0 and \frac{1}{2}e^{-x} for x > 0.

I've found the moment generating function, and deduced that it has mean 0 and variance 2.

However, I'm unable to obtain the moment generating function for \frac{Y}{\sqrt{2n}}. The mark scheme says this:

If T = \frac{Y}{\sqrt{2n}}, then M_{T}(\theta) = E(e^{T\theta}) = E(e^{\theta \sum \frac{X_{i}}{\sqrt{2n}}}) = \prod_{i=1}^{n}E(e^{\frac{\theta}{\sqrt{2n}}X_{i}}) = \left( 1 - \frac{\theta^{2}}{2n} \right)^n

I understand everything up until where the last part; how are they turning that product into that neat (1 - theta^2 / 2n)^n term? My approach was to say that all the X_i have the same distribution, so every term in the product is the same, and you get this:

\left( \frac{1}{\sqrt{2n}}M_{X}(\theta) \right)^n = \left(\frac{1}{\sqrt{2n}(1 - \theta^{2})} \right)^n

but this is clearly not equivalent to their answer. What have I done wrong here and what have they done to collapse their product into something so simple?

Careful: $M_{X/\sqrt{2n}}(\theta) = E \exp(\theta X /\sqrt{2n})= M_X(u), \: u = \theta /\sqrt{2n}.$ This is NOT equal to $(1/\sqrt{2n}) M_X(\theta).$ However, you are partly right: they should not have written $(1- \theta^2/2n)^n$; it should be $(1- \theta^2 / 2n)^{-n}.$

 

[FeDeX_LaTeX]

Careful: $M_{X/\sqrt{2n}}(\theta) = E \exp(\theta X /\sqrt{2n})= M_X(u), \: u = \theta /\sqrt{2n}.$ This is NOT equal to $(1/\sqrt{2n}) M_X(\theta).$ However, you are partly right: they should not have written $(1- \theta^2/2n)^n$; it should be $(1- \theta^2 / 2n)^{-n}.$

Sorry, that was my typo -- they did write that term to the negative power of n.

Okay thanks, I think that makes sense -- so M_{X}(a \theta) \neq aM_{X}(\theta)?

Thanks, it makes so much sense now, they've just replaced the theta with theta / sqrt(2n) :)