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Let the probability density function of a random variable $X$ be given by

$f(x)=\alpha e^{-x^2-\beta x}\ \ \ \ \ \ \infty<x<\infty $

If $E(X)= -\dfrac{1}{2}$ then

$(A) \ \alpha =\dfrac{e^{\frac{-1}{4}}}{\sqrt \pi}$ and $\beta=1$

$(B) \ \alpha =\dfrac{e^{\frac{-1}{4}}}{\sqrt \pi}$ and $\beta=-1$

$(C) \ \alpha =e^\frac{-1}{4}\sqrt \pi$ and $\beta=-1$

$(D) \ \alpha =e^\frac{-1}{4}\sqrt \pi$ and $\beta=1$

The way I did this question:

$\int_{-\infty}^{\infty}f(x)dx=\int_{-\infty}^{\infty}\alpha e^{-(x^2+\beta x +\frac{\beta^2}{4}-\frac{\beta^2}{4})}dx=\int_{-\infty}^{\infty}\alpha e^{-(x+\frac{\beta}{2})^2}e^{\frac{\beta^2}{4}}dx=1 $

put $y=(x+\frac{\beta}{2})$

$\int_{-\infty}^{\infty}\alpha e^{-(y)^2}e^{\frac{\beta^2}{4}}dx=\alpha\sqrt{\pi}e^{\frac{\beta^2}{4}}=1\implies \alpha=\dfrac{e^{\frac{-\beta^2}{4}}}{\sqrt{\pi}}$

Now using $E(X)=\int_{-\infty}^{\infty}x\alpha e^{-(x+\frac{\beta}{2})^2}e^{\frac{\beta^2}{4}}dx$

put $y=(x+\frac{\beta}{2})$

$E(X)=\int_{-\infty}^{\infty}(y-\frac{\beta}{2})\alpha e^{-(y)^2}e^{\frac{\beta^2}{4}}dx=\underbrace{\alpha e^{\frac{\beta^2}{4}} \int_{-\infty}^{\infty}(y) e^{-(y)^2}dx}_{\text {odd function}}-\frac{\beta}{2}\underbrace{\int_{-\infty}^{\infty}\alpha e^{-(y)^2}e^{\frac{-\beta^2}{4}}}_{\text{pdf}}dx$

$0-\dfrac{\beta}{2}=-\dfrac{1}{2} \implies \fbox {$\beta$ =1}$ and $\fbox{$\alpha=\dfrac{e^{\frac{-1}{4}}}{\sqrt{\pi}}$} $

Somehow after doing rigorous calculations, I got this answer but this question came in 2 marks so I am looking for a quick way out. Any alternate solution which less time consuming ?

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$f(x)$ is in the form of a normal PDF: $\frac{1}{\sqrt{2\pi\sigma^2}}\exp(-(x-\mu)^2/2\sigma^2)$, which should be equal to $\alpha\exp(-x^2-\beta x)$. Coefficients should match in LHS and RHS, so $2\sigma^2=1$ and $E[X]=\mu=-1/2$. Then, LHS becomes $$\frac{1}{\sqrt{\pi}}\exp(-x^2+2\mu x - \mu^2)=\alpha\exp(-x^2-\beta x)$$ which gives you $\beta=1$ and $\alpha=\frac{\exp(-1/4)}{\sqrt{\pi}}$.

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  • $\begingroup$ Omg I feel so dumb now . $\endgroup$ – Daman deep Jan 27 at 13:07
  • $\begingroup$ no you're path independent :) $\endgroup$ – gunes Jan 27 at 13:08

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