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I'm trying to wrap my head around linear transformations to random variables (with coefficients > 1).

Consider the two random and independent variables $X$ and $Y$ where:

$$X \sim \mathcal{N}(0,1)\qquad Y \sim \mathcal{N}(-1, 2)$$

I'm trying to determine the distribution of $Z = 2X + 3Y$.


The attempt so far:

I've defined the MGF of both $X$ and $Y$ as:

\begin{align*}M_X(s) &= \exp(s \mu + \sigma^2 s^2 / 2) = \exp(s^2/2)\\ M_Y(s) &= \exp(-s + s^2)\\ \end{align*}

Then the plan was to calculate $U$ as:

$Z = 2X + 3Y = 2 \exp(s^2/2) \cdot 3 \exp(-s + s^2) = 6 \exp(3s^2/2 - s)$

This lead me to a dead end and I'm not sure if this is the right approach, the "new" MGF doesn't exactly look like anything I know.

I've considered using a convolution to try and solve this but I am not 100% about that approach.

There was a mention on this Wikipedia page about how to do it for the case of the coefficients = 1 and that there was a special case for coeffs > 1 (however it failed to expand on it).


I've done some simulations in MATLAB with random sampling from both of these Gaussians and then applying the transformation. The results indicate the solution should be (assuming the code is correct):

$Z \sim \mathcal{N}(-3, 22)$

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  • $\begingroup$ in general, sum of independent random variables are normal. So you just need to compute the mean and variance (immediate) and you have the result $\endgroup$
    – Ant
    Commented May 7, 2018 at 12:24
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    $\begingroup$ I believe it is also worth pointing out the error in the steps that brought the OP to the dead end. Specifically, if $X$ has MGF $M_X(s)$, then $a \, X$ has MGF $M_X(a \, s)$, NOT $a \, M_X(s)$. $\endgroup$ Commented May 13, 2018 at 14:55
  • $\begingroup$ The OP didn't say he wanted it solved using mgfs. He just demonstrated it as his attempt. Christian showed to do it with mgfs. I showed an alternative approach. $\endgroup$ Commented May 13, 2018 at 17:36

3 Answers 3

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First, let me note there is nothing special in having the coefficients of the linear combination to be less or more than one.

The moment generating function is defined as $$M_X(s)=\mathbb{E}[\exp\{sX\}]$$ when this expectation exists. Considering a linear combination of independent random variables, like $2X+3Y$, leads to the moment generating function \begin{align} M_{2X+3Y}(s)&=\mathbb{E}[\exp\{s(2X+3Y)\}]\tag{definition}\\&=\mathbb{E}[\exp\{s2X\}\exp\{s3Y\}]\\&=\mathbb{E}[\exp\{2sX\}]\mathbb{E}[\exp\{3sY\}]\tag{independence}\\&=M_X(2s)M_Y(3s)\tag{identification}\\&=\exp\{4s^2/2\}\exp\{-3s+9s^2\}\tag{normality}\\&=\exp\{22s^2/2-3s\}\end{align} which uniquely and perfectly identifies a ${\cal N}(-3,22)$ distribution. The very same steps can be used to establish that any linear combination of two independent Normal variates is again a Normal variate.

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  • $\begingroup$ In its generalized form, this is the sum of weighted normal random variables, and the proof looks like this (from my grad school notes) suchanutter.net/ItCanBeShown/… $\endgroup$
    – Benjamin
    Commented May 7, 2018 at 13:37
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  1. Use MGF to determine that a linear combination of normal random variables is normal. (MGF uniquely defines the distribution)

  2. Since the sum of normals is normal, take the expectation and variance of $2X + 3Y$ to find the parameters governing the normal distribution. Use the property of variance: $V(\sum X_i) = \sum V(X_i)$ if $X_i$ are independent.

(I'm assuming you made a typo by saying $Z = 2X + 3Z$ and meant $Z = 2X + 3Y$).

Also: Your simulation is correct.

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You don't need to use moment generating functions. The sum of two independent normal random variables is normal with mean equal to the sum of the means and the variance equal to the sum of the variances. Also a constant c times a normal random variable is normal with mean c$\mu$ where $\mu$ is the mean of the original normal and variance equal to $c^2$ $\sigma^2$ where $\sigma^2$ is the variance of the original normal.

Given this $2X$ is normal with mean 0 and variance 4. 3Y is noemal with mean -3 and variance 9(2)=18 Therefore $2X+3Y$ is normal with mean -3 and variance 4+18 =22.

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    $\begingroup$ Resorting to the moment generating functions is however most convenient to establish/prove these properties of the Normal distribution when starting from scratch, e.g., defining the Normal distribution via its mgf. $\endgroup$
    – Xi'an
    Commented May 7, 2018 at 7:28
  • $\begingroup$ If X and Y are independent and normal they are uncorrelated and any linear combination of 2 independent normals is normal as is well known. $\endgroup$ Commented May 9, 2018 at 16:11
  • $\begingroup$ Why do you say that? My argument gets the result only by applying properties of the normal distribution. $\endgroup$ Commented May 9, 2018 at 18:10
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    $\begingroup$ -1. The OP is exactly trying to establish the result using the MGF. Simply saying that it is the property of normal distribution is not helping the original goal. $\endgroup$ Commented May 13, 2018 at 14:51
  • $\begingroup$ To this read my comment above. $\endgroup$ Commented May 13, 2018 at 17:37

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