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Given the following

$Y_1 \sim \mathcal{N}(μ, σ^2 )$

and

$Y_2=α+βY_1+U \;where \; Y_1 \;and \;U\;is\;independent\;and\;U∼\mathcal{N}(0,v^2)$

Let $μ=350$ and $σ^2 =12365$

How would i calculate the expected value and variance from $Y_2$ ? And how can i find the distribution of it?

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3 Answers 3

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Pretty much by definition of expected value: $$E(Y_2) = E(\alpha + \beta Y_1 + U)\\ = E(\alpha) + E(\beta Y_1)+ E(U)\\ = \alpha + \beta E(Y_1) + E(U)$$

And variance: $$var(Y_2) = E(Y_2^2) - E(Y_2)^2$$ We need to find $E(Y_2^2)$, which is

$$E\left[(\alpha + \beta Y_1 + U)^2\right]\\ = E\left[\alpha^2 + \alpha \beta Y_1 + \alpha U\\ + \alpha \beta Y_1 + \beta^2 Y_1^2 + \beta Y_1 U\\ + \alpha U + \beta Y_1 U + U^2\right].$$

The rest you can probably do with linearity and some arithmetic, I hope this helps.

Edit: Most of the work comes from linearity of the expect value function.

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  • $\begingroup$ I updated my post with more information if it makes any difference, cause im not sure i understand $\endgroup$
    – Marcus F
    Commented Mar 20, 2021 at 16:57
  • $\begingroup$ It seems you added the specific values, but that does not change how we approach the problem. We can solve it abstractly, and then plug in the numbers in the end. $\endgroup$
    – JLMF
    Commented Mar 20, 2021 at 17:33
  • $\begingroup$ I see but im not sure how to calculate the expected value and variance using the formulas you gave. $\endgroup$
    – Marcus F
    Commented Mar 20, 2021 at 17:55
  • $\begingroup$ What is the distribution in this case @BruceET $\endgroup$
    – Marcus F
    Commented Mar 21, 2021 at 4:47
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Because of independence:

$m_Y=E(Y_2)=\alpha +\beta \mu$ $\sigma^2_Y=var(Y_2)=var(\alpha)+var(\beta Y_1)+var (U)$

$=\beta \sigma^2+ v^2$

$Y_2$ ~ $N(m_Y,\sigma^2_Y)$.

You need values for $\alpha$, $\beta$ and $v^2$.

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  • $\begingroup$ I think alpha and beta can just be variables and not concrete values $\endgroup$
    – Marcus F
    Commented Mar 20, 2021 at 21:17
  • $\begingroup$ In that case you can't get a numerical result for mean or variance. $\endgroup$ Commented Mar 20, 2021 at 21:39
  • $\begingroup$ So what would the expected value and variance be in this case without concrete values? Sorry im not that good at this $\endgroup$
    – Marcus F
    Commented Mar 20, 2021 at 21:43
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    $\begingroup$ The formulas given in the answer are all you can get. $\endgroup$ Commented Mar 20, 2021 at 21:47
  • $\begingroup$ If i have to split it up into $E(Y_2) = something$ and $Var(Y_2) = something$ how would i do it from the formulas you've given? and i guess $Y_2$ ~ $𝑁(𝑚_𝑌,𝜎^2_Y)$ is the distribution but what is $m_Y$ ? and can it be written in another way? $\endgroup$
    – Marcus F
    Commented Mar 20, 2021 at 22:31
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You still haven't given values for $v, a, b,$ so I have supplied some.

The distribution of $Y_2$ is normal.

In my simulation in R below, I have used a million iterations, so the sample mean and SD of $Y_2$ should agree with the respective population values to several significant digits--I guess, good enough to find mistakes in applying the formulas.

set.seed(2021)
y1 = rnorm(10^6, 350, 111.198)
u =  rnorm(10^6, 0, 50)
a = 10;  b = 50
y2 = a + b*y1 + u
summary(y2);  sd(y2);  var(y2)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  -8248   13765   17511   17513   21263   43899 
[1] 5566.788  # aprx SD(Y2)
[1] 30989134  # aprx Var(Y2)

Here are histograms of the simulated distributions along with the corresponding normal density curves.

enter image description here

R code for figure.

par(mfrow=c(1,3))
hist(y1, prob=T, col="wheat")
 curve(dnorm(x,350,111.198), add=T, lwd=2)
hist(u, prob=T, col="wheat")
 curve(dnorm(x,0,50), add=T, lwd=2)
hist(y2, prob=T, col="skyblue2")
 curve(dnorm(x, 17813, 5566.788), add=T, lwd=2)
par(mfrow=c(1,1))
  • You say in a comment that you don't follow how to calculate some of the results for $Y_2.$ That is not a comment that prompts specific explanations.

  • I have had a quick look at answers by @herbsteinberg and @JLF, which seem fine: (+1) for each.

  • Can you say specifically what you don't understand? If there are several difficulties, you might start by saying what the first couple of them are.

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  • $\begingroup$ In @JLF 's answer im not sure what the distribution is $\endgroup$
    – Marcus F
    Commented Mar 21, 2021 at 4:49
  • $\begingroup$ Do you mean the distribution of $Y_2?$ That's also normal. The purpose is to find its mean and variance. $\endgroup$
    – BruceET
    Commented Mar 21, 2021 at 5:48
  • $\begingroup$ I mean in terms of notation, how would it be written, if im following @JlF 's advice? $ X \sim something$ also is it possible to find the PDF? Probability density function? $\endgroup$
    – Marcus F
    Commented Mar 21, 2021 at 17:47
  • $\begingroup$ $Y_2 \sim \mathsf{Norm}(\mu, \sigma),$ where my simulation shows $\mu\approx 17513, \sigma\approx 5566.788,$ and you can get exact values of $\mu$ and $\sigma$ using the other answers. Once $\mu$ and $\sigma$ are known, plug them into the general formula for a normal density function. (This density function is not a secret; most prob and stat books have it.) $\endgroup$
    – BruceET
    Commented Mar 21, 2021 at 19:04

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