Moment Generating Function (1 Viewer)

integral95 · May 31, 2016

$$ For any $ \ \ 0<p<1 \ \ $and$ \ \ r \ \ $ a positive integer, the probability function $ \\ \\ f(x) = \binom{r+x+1}{x}p^r(1-p)^x \ \ \ x = 0,1,2 ... \\ \\ $defines a random variable X$ \\ \\ $Show that the moment generating function$ \ \ m_X(u) = \mathbb{E}(e^{ux}) = \\ (\frac{p}{1-(1-p)e^u})^r \\ \\ \\ \\ $Define a new variable by $ Y = 2pX \\ \\ $Write down the moment generating function$ \ \ M_Y(u) \ \ $of$ \ \ Y$

Yeah like I know how the formula works, but the algebra seems pretty intense I don't really know where to go.

eyeseeyou · May 31, 2016

integral95 said:
$$ For any $ \ \ 0<p<1 \ \ $and$ \ \ r \ \ $ a positive integer, the probability function $ \\ \\ f(x) = \binom{r+x+1}{x}p^r(1-p)^x \ \ \ x = 0,1,2 ... \\ \\ $defines a random variable X$ \\ \\ $Show that the moment generating function$ \ \ m_X(u) = \mathbb{E}(e^{ux}) = \\ (\frac{p}{1-(1-p)e^u})^r \\ \\ \\ \\ $Define a new variable by $ Y = 2pX \\ \\ $Write down the moment generating function$ \ \ M_Y(u) \ \ $of$ \ \ Y$

Yeah like I know how the formula works, but the algebra seems pretty intense I don't really know where to go.

Is this a HSC question?

integral95 · May 31, 2016

photastic · May 31, 2016

Looks like MATH2901.

integral95 · May 31, 2016

photastic said:
Looks like MATH2901.

Dammit, who are you ?

Bump... pls help

seanieg89 · May 31, 2016

integral95 said:
$$ For any $ \ \ 0<p<1 \ \ $and$ \ \ r \ \ $ a positive integer, the probability function $ \\ \\ f(x) = \binom{r+x+1}{x}p^r(1-p)^x \ \ \ x = 0,1,2 ... \\ \\ $defines a random variable X$ \\ \\ $Show that the moment generating function$ \ \ m_X(u) = \mathbb{E}(e^{ux}) = \\ (\frac{p}{1-(1-p)e^u})^r \\ \\ \\ \\ $Define a new variable by $ Y = 2pX \\ \\ $Write down the moment generating function$ \ \ M_Y(u) \ \ $of$ \ \ Y$

Yeah like I know how the formula works, but the algebra seems pretty intense I don't really know where to go.

Try to exploit the fact that you are given that that binomial expression is a pdf, ie it sums to 1 for any value of the parameter p. (*)

Try to evaluate the sum that occurs in E(e^(ux)) by manipulating it to contain an expression that you can sum using (*)

For the last bit, again try to manipulate it so you can kill the hardest part of the summation with (*).

The algebra is not difficult without needing to prove (*) yourself, but doing so is a good exercise.

InteGrand · May 31, 2016

integral95 said:
$$ For any $ \ \ 0<p<1 \ \ $and$ \ \ r \ \ $ a positive integer, the probability function $ \\ \\ f(x) = \binom{r+x+1}{x}p^r(1-p)^x \ \ \ x = 0,1,2 ... \\ \\ $defines a random variable X$ \\ \\ $Show that the moment generating function$ \ \ m_X(u) = \mathbb{E}(e^{ux}) = \\ (\frac{p}{1-(1-p)e^u})^r \\ \\ \\ \\ $Define a new variable by $ Y = 2pX \\ \\ $Write down the moment generating function$ \ \ M_Y(u) \ \ $of$ \ \ Y$

Yeah like I know how the formula works, but the algebra seems pretty intense I don't really know where to go.

$\noindent By the way, once you've proven the formula for $m_{X}\left(u\right)$ (or even if you don't manage to prove it), the one for $Y$ is easy (since the one for $X$ is given), it's just $m_{Y} \left(u\right)= m_{X} \left(2pu\right)$. In general $m_{aX}\left(u\right)$ will be just $m_{X} \left(au\right)$ for any random variable $X$ with an MGF and $a$ being a constant.$

Moment Generating Function (1 Viewer)

integral95

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eyeseeyou

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integral95

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photastic

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integral95

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seanieg89

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InteGrand

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